I am having a problem determining the thermal conductivity of a layered sample. The sample is made of two materials, one whose thermal conductivity is known (λ-1) and one whose thermal conductivity is unknown (λ-2).

The thermal conductivity of the entire sample is known by means of testing. I am trying to calculate λ-2, and my first thought was to use the one-dimensional steady state diffusion equation without heat generation:

$$ \frac{d}{dx}(\lambda \frac{dT}{dx}) = 0$$

Next, I will enter the known and unknown values:

$$ \frac{d}{dx_{total}}(\lambda_{total} \frac{dT}{dx_{total}}) = \frac{d}{dx_{1}}(\lambda_{1} \frac{dT}{dx_{1}}) + \frac{d}{dx_{2}}(\lambda_{2} \frac{dT}{dx_{2}}) $$

Is this approach correct? I do not know dT (perhaps can find in machine's documentation), I'm not sure if the addition on the right side of the equation is legit; am I headed in the wrong direction?

The tested layered structure

A simplified model for analysis


1 Answer 1


The thermal resistance for a slab in steady-state heat transfer is $R_t = w/(kA)$ (K/W), where $w$ is thickness (m), $k$ thermal conductivity (W/m K), and $A$ cross-sectional area (m$^2$). For serial thermal resistors as your case

$$R_t = \sum R_{t,j} $$

For $N_j$ layers at each $k_j$ thermal conductivity and $w_j$ width, this gives

$$R_t = \frac{N_1\ w_1 + N_2\ w_2}{k_t} = \frac{N_1\ w_1}{k_1}+ \frac{N_2\ w_2}{k_2}$$

and with $w_1 = w_2$ and $N_1 = N_2 = N$ we find

$$R_t = \frac{2}{k_t} = \frac{1}{k_1} + \frac{1}{k_2}$$

  • $\begingroup$ Thank you @Jeffrey. This is related to a side project we are working on in the office, your post is a huge help. $\endgroup$
    – codenoob
    Commented Apr 13, 2022 at 14:00
  • $\begingroup$ @codenoob FWIW, as a new contributor, the next acceptable step is to (up/down)vote or to mark as complete. $\endgroup$ Commented Apr 14, 2022 at 0:29
  • $\begingroup$ Got it - thanks $\endgroup$
    – codenoob
    Commented Jun 10, 2022 at 15:22

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