Thermal resistance, can i do this?

In summary: The wall has three layers - fiberboard on the inside, concrete in the middle, and brick on the outside. Each layer has a different thickness and thermal conductivity. Jessie knows that thermal resistance is calculated by dividing the thickness by the conductivity constant, and decides to combine the resistances by simply adding them together. After rearranging the formula, Jessie gets a k value for the entire wall and uses it to find the amount of energy that will conduct through the wall.
  • #1
ShawnD
Science Advisor
718
2
I have this problem from my book here
A house has a wall made of 3 layers: fiberboard, concrete, brick where the fiberboard is on the inside of the house, brick is on the outside and concrete is in between.
the fiberboard is 0.02m thick, the concrete is 0.15m thick, the brick is 0.07m thick.
The thermal conductivity for fiberboard is 0.059, the concrete is 1.3, the brick is 0.71
the inside temperature is 20C, the outside temp is -10C
I'm supposed to find the amount of energy to conduct through the wall in 1 hour.

Since I have too many layers, I couldn't do the method of finding out temperature between the layers unless I made a pile of equations then put them in a matrix (which I hate doing).
I got the idea that I should try to make 1 big value for the conductivity of the wall.

Thermal resistance is given by this equation:
R = d/k where d is thickness and k is conductivity constant

I combined the resistances by simply adding them together (can I do this??)
[sum]R = d/k + d/k + d/k
[sum]R = 0.02/0.059 + 0.15/1.3 + 0.07/0.71
[sum]R = 0.504067

Then I rearrange the formula
R = d/k
Rk = d
k = d/R
k = (0.02 + 0.15 + 0.07)/0.504067
k = 0.476127 for the entire wall

Now that I got a k value for the entire wall, I just fill in for
Q = kA[del]Tt/d



Can I do the stuff I did with the adding of resistances like that or did I just do a bunch of voodoo math that is totally wrong?
 
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  • #2
Try

http://personal.cityu.edu.hk/~bsapplec/newpage218.htm

It seems to answer some bits of your question. From what I've found, it looks like you can combine thermal resistances in series the way you've done it.

Jess
 
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  • #3


Yes, you can combine thermal resistances in series by simply adding them together. This is a valid method and is commonly used in thermal calculations. In this case, you correctly found the total thermal resistance of the wall by adding the individual resistances. Your calculation for the overall thermal conductivity of the wall is also correct. Great job! Just make sure to pay attention to units and use the correct values in your final calculation for the amount of energy conducted through the wall in 1 hour.
 

1. What is thermal resistance and why is it important?

Thermal resistance refers to the measure of how difficult it is for heat to flow through a material. It is an important concept in the study of heat transfer and is used to determine the effectiveness of insulation materials, as well as the efficiency of various systems such as heating and cooling systems.

2. How is thermal resistance calculated?

Thermal resistance is calculated by dividing the thickness of a material by its thermal conductivity. It is represented by the symbol R and is measured in units of square meters kelvin per watt (m²K/W).

3. What factors affect thermal resistance?

The thermal resistance of a material is affected by its thickness, thermal conductivity, and surface area. Thicker materials and those with higher thermal conductivity will have lower thermal resistance, while a larger surface area can increase thermal resistance.

4. Can I improve thermal resistance?

Yes, thermal resistance can be improved by using materials with lower thermal conductivity, increasing the thickness of the material, or adding insulation layers. Additionally, reducing the surface area of the material can also help improve thermal resistance.

5. How is thermal resistance used in real-world applications?

Thermal resistance is used in a variety of real-world applications, including building insulation, electronics cooling, and thermal clothing. It is also an important factor in the design and efficiency of heating and cooling systems in homes and buildings.

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