# Calculate Thermal Conductivity of Layered Structure

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 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

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

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