Thermal resistance, can i do this?

[ShawnD]

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?

 

[Jess]

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

 

[Graeme Robinson]

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.