Conduction with Variable Conductivity k = f(T)
Solve steady conduction inside geometries with variable thermal conductivity using the Kirchhoff transformation.
Calculation Domain Inputs
Analyze conduction through solid bodies where thermal conductivity changes with local temperature:
- Geometry Selection: Evaluate flat walls, cylinders, or spherical shells.
- k(T) Model: Choose between linear, polynomial, or tabulated conductivity functions.
- Kirchhoff transformation: Converts non-linear equations to linear form to compute exact heat transfer rates and profiles.
- Constant-k comparison: Visualize the non-linear temperature gradient vs. the standard linear/ideal case.
Geometry & Property Setup
$$\theta(T) = \int_0^T k(T') dT'$$ $$Q = C_{geom} (\theta_1 - \theta_2)$$ • Plane Wall: $C_{geom} = A / L$
• Cylinder: $C_{geom} = 2\pi H / \ln(r_2/r_1)$
• Sphere: $C_{geom} = 4\pi r_1 r_2 / (r_2 - r_1)$
Results & Visualization
Temperature Profile Overlay
Thermal Conductivity Variation
Detailed Spatial Analysis
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Calculation Methodology
Mathematical Model & Theory
Steady-state 1D heat conduction with variable thermal conductivity $k(T)$ is governed by:
We solve this equation by introducing the Kirchhoff Transformation:
This linearizes the conduction equation into Laplace's equation $\nabla^2 \theta = 0$. Once solved, the thermal profile of $\theta$ is mapped back to $T$ using the inverse function $T = \theta^{-1}(\theta)$.
Geometry Integrals & Heat Transfer (Q):
- Plane Wall: $Q = \frac{A}{L}(\theta_1 - \theta_2)$
- Cylinder: $Q = \frac{2\pi H}{\ln(r_2/r_1)}(\theta_1 - \theta_2)$
- Sphere: $Q = \frac{4\pi r_1 r_2}{r_2 - r_1}(\theta_1 - \theta_2)$
Academic References:
- Kirchhoff, G. (1894). Vorlesungen über die Theorie der Wärme.
- Incropera, F. P., & DeWitt, D. P. (2011). Fundamentals of Heat and Mass Transfer.
Worked Engineering Example
A plane wall of thickness $L = 0.1$ m, area $A = 1$ m² has linear variable conductivity: $$k(T) = 1.5 (1 + 0.003 T) \text{ W/(m·K)}$$ Boundary temperatures are $T_1 = 300^\circ\text{C}$ and $T_2 = 50^\circ\text{C}$. Find the heat transfer rate $Q$ and compare it with the constant-k average case.
Step-by-step Solution:
1. Evaluate $\theta(T) = 1.5 (T + 0.0015 T^2)$ at boundaries:
$$\theta_1 = 1.5 (300 + 0.0015 \times 90000) = 652.5 \text{ W/m}$$ $$\theta_2 = 1.5 (50 + 0.0015 \times 2500) = 80.625 \text{ W/m}$$ 2. Calculate Heat Transfer rate $Q$ via Kirchhoff:
$$Q = \frac{A}{L} (\theta_1 - \theta_2) = \frac{1}{0.1} (652.5 - 80.625) = 5718.75 \text{ W}$$ 3. Evaluate constant-k case at average temperature $T_{avg} = 175^\circ\text{C}$:
$$k_{avg} = 1.5 (1 + 0.003 \times 175) = 2.2875 \text{ W/(m·K)}$$ $$Q_{const} = \frac{k_{avg} A}{L}(T_1 - T_2) = \frac{2.2875 \times 1}{0.1}(300 - 50) = 5718.75 \text{ W}$$ Note that for linear properties, evaluating constant-$k$ at the arithmetic mean temperature yields the exact same heat flow rate, but the spatial temperature profile inside the wall is non-linear!