THERMAL CYU 11

1. Define Conductivity and Resistivity:

The problem involves anisotropic conduction where thermal conductivity depends on direction. Given:

• Conductivity along the fibres (parallel): $K_{\parallel} = \eta K$
• Conductivity perpendicular to fibres: $K_{\perp} = K$

In cases where the current direction is constrained (by insulated boundaries), it is easier to work with Thermal Resistivity ($\rho$), which is the reciprocal of conductivity.

• Resistivity along fibres: $\rho_{\parallel} = \frac{1}{K_{\parallel}} = \frac{1}{\eta K}$
• Resistivity perpendicular to fibres: $\rho_{\perp} = \frac{1}{K_{\perp}} = \frac{1}{K}$

2. Cylinder 1: Axis Parallel to Fibres

The cylinder axis is aligned with the fibres. The heat flows directly along the path of lowest resistivity. $$ \rho_{1} = \rho_{\parallel} = \frac{1}{\eta K} $$ The Heat Flux $J_1$ for a temperature gradient $\frac{\Delta T}{L}$ is: $$ J_1 = \frac{1}{\rho_1} \cdot \frac{\Delta T}{L} = \eta K \frac{\Delta T}{L} $$

3. Cylinder 2: Axis at Angle $\theta$

Constraint: The curved surfaces are “well insulated.” This means no heat can flow out of the sides ($\vec{J} \cdot \hat{n} = 0$). Consequently, the Heat Flux vector $\vec{J}$ is forced to be strictly along the axis of the cylinder.

To find the flux, we calculate the effective resistivity along the cylinder axis ($\rho_{eff}$) by projecting the resistivity tensor onto the axis direction: $$ \rho_{eff} = \rho_{\parallel} \cos^2 \theta + \rho_{\perp} \sin^2 \theta $$ Substituting the values: $$ \rho_{eff} = \left( \frac{1}{\eta K} \right) \cos^2 \theta + \left( \frac{1}{K} \right) \sin^2 \theta $$ $$ \rho_{eff} = \frac{\cos^2 \theta + \eta \sin^2 \theta}{\eta K} $$ The effective conductivity $K_{eff}$ is the inverse of $\rho_{eff}$: $$ K_{eff} = \frac{1}{\rho_{eff}} = \frac{\eta K}{\cos^2 \theta + \eta \sin^2 \theta} $$ Thus, the Heat Flux $J_2$ is: $$ J_2 = K_{eff} \frac{\Delta T}{L} = \left( \frac{\eta K}{\cos^2 \theta + \eta \sin^2 \theta} \right) \frac{\Delta T}{L} $$

4. Calculate the Ratio

We need the ratio of heat flux in the first cylinder to that in the second cylinder ($J_1 / J_2$). $$ \text{Ratio} = \frac{J_1}{J_2} = \frac{ \eta K \left( \frac{\Delta T}{L} \right) }{ \left[ \frac{\eta K}{\cos^2 \theta + \eta \sin^2 \theta} \right] \left( \frac{\Delta T}{L} \right) } $$ Simplifying the expression: $$ \text{Ratio} = \frac{ \eta K }{ \frac{\eta K}{\cos^2 \theta + \eta \sin^2 \theta} } $$ $$ \text{Ratio} = \cos^2 \theta + \eta \sin^2 \theta $$