📈 Variable Conductivity k = f(T)

Solve steady heat conduction through plane walls, cylinders, and spheres with temperature-dependent conductivity k(T) using the Kirchhoff transformation.

⚡ Fortran 90 Engine Double Precision (IEEE 754) ✓ ISO / ASME Validated

Heat Transfer

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📅 Released Jun 2026

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Geometry & Property Setup

Geometry:

Thickness ($L$) [m]:

Surface Area ($A$) [m²]:

k(T) Property Model:

Ref Conductivity ($k_0$) [W/m·K]:

Coefficient ($\beta$) [1/K]:

Surface Boundaries

Inner Temp ($T_1$) [°C]:

Outer Temp ($T_2$) [°C]:

Kirchhoff Transform Formula:
$$\theta(T) = \int_0^T k(T') dT'$$ $$Q = C_{geom} (\theta_1 - \theta_2)$$ • 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

Heat Rate (Q)

Watts

Mean Effective k

W/(m·K)

Constant k_avg Q

Watts

Diff in Heat Loss

Watts

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:

$$\nabla \cdot (k(T) \nabla T) = 0$$

We solve this equation by introducing the Kirchhoff Transformation:

$$\theta(T) = \int_{0}^{T} k(T') dT'$$

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

Problem Statement:
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!