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Engineering Calculation Report
Date: 2026-07-01 02:57

Anisotropic Thermal Conduction

Analyze directional thermal conductivity tensor rotation for composites, PCBs, and wood.

x y k_x (fiber axis) k_y θ Heat Flux Vector (q)

Anisotropic Conductivity Tensor

For non-isotropic materials (composite fibers, wood grains, or printed circuit boards), thermal conductivity varies by direction:

  • $k_x, k_y, k_z$: Principal conductivities along material axes (e.g. parallel & perpendicular to fibers).
  • Rotation ($\theta$): Angle of orientation between principal material coordinates and the thermal system coordinate axis.
  • Rotated $k_{eff}$: Directional thermal conductivity resulting from the coordinate system transformation.
CFRP Composite (30°) PCB Board (45°) Wood (Oak, 15°) Pyrolytic Graphite (60°)

Parameters Input

Directional Tensor Rotation: $$k_{eff}(\theta) = k_x\cos^2\theta + k_y\sin^2\theta$$ $$q_x = k_{eff}\frac{T_{hot} - T_{cold}}{L_x}$$ $$Q_x = q_x A$$

Results & Analysis

Results and visualizations will be displayed here upon completion of the computation.

Calculation Methodology

Mathematical Model & Theory

Fourier's Law for heat conduction in an isotropic solid links heat flux directly to temperature gradient. However, in anisotropic materials (like composites, wood, or graphitic crystals), conductivity is a tensor quantity. The directional heat flux in coordinate systems rotated relative to principal material coordinates in 2D is resolved as:

$$k_{eff}(\theta) = k_x \cos^2(\theta) + k_y \sin^2(\theta)$$ $$k_{xy}' = (k_x - k_y)\sin(\theta)\cos(\theta)$$

Variable Definitions & Units:

  • $k_x, k_y, k_z$: Thermal conductivity along principal axes [W/m·K]
  • $\theta$: Fiber or layer rotation angle [rad or deg]
  • $k_{eff}$: Rotated effective thermal conductivity [W/m·K]
  • $q_x$: Thermal heat flux along x-axis [W/m²]
  • $Q_x$: Heat transfer rate [W]

Academic References:

  1. Incropera, F. P., & DeWitt, D. P. (2011). Fundamentals of Heat and Mass Transfer. Wiley.
  2. Özisik, M. N. (1993). Heat Conduction. Wiley.
  3. Carslaw, H. S., & Jaeger, J. C. (1959). Conduction of Heat in Solids. Oxford University Press.

Worked Engineering Example

Problem Statement:
A Carbon Fiber Reinforced Polymer (CFRP) sheet ($k_x = 7.0$ W/m·K along fibers, $k_y = 0.8$ W/m·K perpendicular to fibers) is 3 mm thick ($L_x = 0.003$ m) and has an area of 0.001 m². The hot face is at 120°C and the cold face is at 25°C. If the fibers are rotated by 30° relative to the thermal system, calculate the effective conductivity and heat rate.

Step-by-step Solution:
1. Calculate the effective rotated conductivity $k_{eff}$:
$$k_{eff} = 7.0\cos^2(30^\circ) + 0.8\sin^2(30^\circ)$$ $$k_{eff} = 7.0(0.75) + 0.8(0.25) = 5.25 + 0.20 = 5.45 \text{ W/m·K}$$ 2. Calculate directional thermal heat flux $q_x$:
$$q_x = k_{eff} \frac{T_{hot} - T_{cold}}{L_x} = 5.45 \frac{120 - 25}{0.003} = 172,583.3 \text{ W/m}^2$$ 3. Calculate heat transfer rate $Q_x$:
$$Q_x = q_x A = 172,583.3 \times 0.001 = 172.58 \text{ W}$$
Final Result:
Effective rotated conductivity is 5.45 W/m·K and total heat rate is 172.6 W.