---
title: "Plane wall with temperature dependent conductivity"
subject: "Heat Transfer"
chapter: 2
key: "semester_5/heat_transfer:variable-thermal-conductivity-plane-wall"
source: https://www.bctengineer.com/q/semester_5/heat_transfer/variable-thermal-conductivity-plane-wall
site: BCT Engineering
---

# Plane wall with temperature dependent conductivity

Heat Transfer · Chapter 2: Conduction Heat Transfer

- Asked in: year not given, 6 marks
- Video: [Tutorial Question 4: Conductive Heat Transfer Through a Plane Wall with Varying Conductivity](https://www.youtube.com/watch?v=dVH2o1PZtes) — Smartmodium
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/heat_transfer/variable-thermal-conductivity-plane-wall
- Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-2

## Question

The thermal conductivity of a material varies with temperature as k = k0(1 + bT). Derive an expression for the steady heat flow through a plane wall of thickness L and area A with surface temperatures T1 and T2, and show that the temperature distribution is non-linear. Find the heat flux through a 0.25 m thick wall with k0 = 0.8 W/m K, b = 0.0006 per K, T1 = 800 °C and T2 = 100 °C.

## Short answer

Integrating Fourier's law with k = k0(1+bT) gives heat flow using the mean conductivity k_m = k0[1 + b(T1+T2)/2]. For the given wall, k_m = 1.016 W/m K and heat flux q = 2.85 kW/m^2.

## Answer

#### Derivation
For steady one-dimensional flow, $Q$ is constant along $x$:

$$
Q = -kA\frac{dT}{dx} = -k_0(1+bT)A\frac{dT}{dx}
$$

Separate the variables and integrate from $x = 0$ ($T_1$) to $x = L$ ($T_2$):

$$
Q\int_0^L dx = -k_0A\int_{T_1}^{T_2}(1+bT)\,dT
$$

$$
QL = k_0A\left[(T_1 - T_2)+\frac{b}{2}\left(T_1^2-T_2^2\right)\right]
$$

$$
Q = \frac{k_0A}{L}\left[1+\frac{b}{2}(T_1+T_2)\right](T_1-T_2) = \frac{k_mA\,(T_1-T_2)}{L}
$$

So the heat flow is the same as for constant conductivity equal to the **mean value** $k_m = k_0\left[1+b\,\dfrac{T_1+T_2}{2}\right]$, the value of $k$ at the average temperature.

#### Shape of temperature profile
Integrating only up to a general $x$ gives a quadratic in $T$:
$T + \dfrac{b}{2}T^2 = T_1+\dfrac{b}{2}T_1^2 - \dfrac{Q x}{k_0A}$. Since it is quadratic in $T$, $T$ is not linear in $x$. Where $k$ is large the gradient is small, and where $k$ is small the gradient is steep. For $b > 0$ the hot side has the larger $k$, so the profile is flatter near the hot face and steeper near the cold face (concave downward).

#### Numerical value

$$
\begin{aligned}
k_m &= 0.8\left[1 + 0.0006\times\frac{800+100}{2}\right] = 0.8(1.27) = 1.016\ \text{W/m K}\\
q &= \frac{k_m\,(T_1-T_2)}{L} = \frac{1.016\times700}{0.25} = 2845\ \text{W/m}^2
\end{aligned}
$$

**Answer:** $q \approx 2.85\ \text{kW/m}^2$ with $k_m = 1.016$ W/m K.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is heat flow derived when k = k0(1 + bT)?
   A. Use k0 at T1 only
   B. Assume T is linear and skip integration
   C. Separate variables and integrate Q dx = -k0 A (1+bT) dT
   D. Differentiate k with respect to T
   Answer: C. Q is constant, so separate variables and integrate between the surfaces. (Key point: Method)

2. Heat flow equals constant-k flow with k evaluated at:
   A. T1
   B. T2
   C. 0 C
   D. The average temperature (T1+T2)/2
   Answer: D. km = k0[1 + b(T1+T2)/2]. (Key point: Mean conductivity)

3. Why is the temperature profile non-linear?
   A. T + bT^2/2 is linear in x, so T is quadratic-root in x
   B. k is constant
   C. Q varies along x
   D. Generation is present
   Answer: A. The relation is quadratic in T, so T is not linear in x. (Key point: Profile shape)

4. With these data, q is about:
   A. 2240 W/m^2
   B. 2845 W/m^2
   C. 1016 W/m^2
   D. 700 W/m^2
   Answer: B. km = 1.016; q = 1.016 x 700/0.25 = 2845 W/m^2. (Key point: Heat flux)

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