# Finding effective thermal conductivity multi-layer cylinder

How does one find the effective thermal conductivity of a cylindrical body made of multiple layers? E.g. Copper core (k1), insulator 1 (k2) and insulator 2 (k3). I need the effective properties for performing lump calculations.

An example can be seen below. I appreciate any kind of help!

• Which axis is the cylinder conducting through: radial, or axial? – CR Drost Jul 10 '15 at 16:42
• It would be nice for have the formula for both. – Physther Jul 10 '15 at 17:17

## Axial conduction, along the cylinder

Let's first talk about electricity. We know that effective resistance of a parallel portion of a circuit found by the relation: $$\frac{1}{R} = \sum \frac{1}{R_i}$$If you just figure out the relation between conductance and resistance, $$G = \frac{1}{R}$$So $G = \sum G_i$. In fact, conductance in parallel is additive, just add each one up and you'll get the total conductance.

Now about electrical conductivity. $$G = \sigma \frac{A}{l}$$ If you have several conductor materials of the same length, then the total conductance would be $$G = \sum G_i = \sum \sigma_i\frac{A_i}{l} = \frac{1}{l} \sum \sigma_iA_i$$ For the effective "total" conductivity, $$\sigma = G \frac{l}{A} = \frac{l}{A} \cdot\frac{1}{l}\sum \sigma_iA_i = \frac{\sum \sigma_iA_i}{A}$$ Since we know thermal conduction and electrical conduction is analogous, this yields: $$k = \frac{\sum k_iA_i}{A_{total}}$$ or in words:
The effective conductivity of a multi-layer composite material is the weighted mean of each component layer's conductivity, where the weight is the cross-sectional area of each layer.

This formula should work for any composite material, be it radially, vertically or block-layered, as long as each cross-section is identically structured. As a demonstration with your question's examples, the effective conductivity would be: $$k = \frac{k_1A_1 + k_2A_2 + k_3A_3}{A_{total}}$$

Note: there is a possibility that the formula even work if each cross-section is not identically structured, as long as the cross-sectional area of each component remains the same, though I didn't bother to proof that.

This one would involve calculus. Starting from the resistance of each layer, $$R_i = \int^{r_i}_{r_{i-1}} \frac{1}{k_iA}dr = \int^{r_i}_{r_{i-1}} \frac{1}{2k_i\pi rl}dr = [\frac{\ln r}{2k_i\pi l}]^{r_i}_{r_{i-1}} = \frac{\ln \frac{r_i}{r_{i-1}}}{2k_i\pi l}$$ Evidently this formula doesn't allow a point source of heat, so you'll need to assume that temperature is homogeneous in any cross-section of the copper component, and then find the resistance of the outer two resistor layers.

In order to solve this, we must first find the total thermal resistance ($R_{total}$). The thermal resistors (copper, insulator 1, and insulator 2) are acting in parallel. Thus the total resistance is give by the following equation:

${\frac{1}{R_{total}}}={\frac{1}{R_1}}+{\frac{1}{R_2}}+{\frac{1}{R_3}}+...+{\frac{1}{R_n}}$ where ${R_n}={\frac{L}{k_nA_n}}$ (this is very similar to parallel resistors in an electric circuit)

${R_n}$ is the thermal resistance of each layer, $L$ is the legnth of the cylinder, ${k_n}$ is the thermal conductivity of each layer, and ${A_n}$ is the cross section area of each layer. When we plug in $k_1,k_2,k_3,A_1,etc.$ we find that ${R_{total}}=\frac{L}{k_1A_1+k_2A_2+k_3A_3+...+k_nA_n}$

If we want to find the effective thermal conductivity (${k_{eff}}$), we know that $R_{total}=\frac{L}{k_{eff}A_{total}}$

Thus $k_{eff} = \frac{L}{R_{total}A_{total}}=\frac{L(k_1A_1+k_2A_2+...+k_nA_n)}{L(A_1+A_2+...+A_n)}=\frac{(k_1A_1+k_2A_2+...+k_nA_n)}{(A_1+A_2+...+A_n)}$