Solving Heat Transfer with Variable Conductivity

Hello! Welcome to our next lesson.

In our previous lessons, we've built a solid foundation for analyzing heat conduction, assuming that thermal conductivity, $k$ , is a constant property of the material. We derived expressions for temperature profiles and heat flow in various geometries, including composite systems.

However, in many real-world materials and certainly in problems designed to test deep understanding (like in the JEE Advanced), thermal conductivity isn't constant—it changes with temperature.

Today, we will tackle exactly that scenario. Our learning outcome is to tackle problems with temperature-dependent thermal conductivity. You'll find that your strong background in calculus is the key to mastering this topic. The fundamental physics (Fourier's Law) remains the same; we just need to handle the mathematics more carefully.

1. The General Approach: Back to Calculus

Let's start with the differential form of Fourier's Law for one-dimensional steady-state heat conduction without heat generation:

$H = - k A \frac{d T}{d x}$

Here, $H$ is the heat current (rate of heat flow, in Watts). In steady state, $H$ must be constant throughout the material. Our crucial change is that we now consider $k$ as a function of temperature, $k (T)$ .

Since $k$ depends on $T$ , and $T$ depends on position $x$ , we can't just integrate $\frac{d T}{d x}$ trivially anymore. The correct approach is to separate the variables before integrating. We group all terms dependent on position on one side and all terms dependent on temperature on the other.

$H d x = - k (T) A d T$

Assuming the area $A$ is constant (like in a plane wall), we can rearrange and set up our definite integral:

$\frac{H}{A} \int_{x_{1}}^{x_{2}} d x = - \int_{T_{1}}^{T_{2}} k (T) d T$

Let's say the wall extends from $x = 0$ (at temperature $T_{1}$ ) to $x = L$ (at temperature $T_{2}$ ). Integrating gives:

$\frac{H L}{A} = \int_{T_{2}}^{T_{1}} k (T) d T$

This gives us the general expression for the heat current:

$H = \frac{A}{L} \int_{T_{2}}^{T_{1}} k (T) d T$

This integral is the core of every problem of this type. Whatever the function $k (T)$ is, we just need to evaluate this integral.

2. The Concept of Average Thermal Conductivity ( $k_{a v g}$ )

Looking at our new formula and comparing it to the old one, $H = \frac{k A ( T _{1} - T _{2} )}{L}$ , we can define an effective or average thermal conductivity, $k_{a v g}$ , such that:

$k_{a v g} (T_{1} - T_{2}) = \int_{T_{2}}^{T_{1}} k (T) d T$

This leads to the formal definition:

$k_{a v g} = \frac{1}{T _{1} - T _{2}} \int_{T_{2}}^{T_{1}} k (T) d T$

This is a powerful concept. It means you can solve a variable-conductivity problem by first calculating the correct average conductivity and then substituting it into the familiar constant-conductivity formula.

Let's explore this further with a reading from our textbook.

HEAT CONDUCTION EQUATION

The chapter 'HEAT CONDUCTION EQUATION' provides a concise explanation of this topic. It introduces the concept of average thermal conductivity and discusses the most common case where conductivity varies linearly with temperature.

Please read section 2–7 VARIABLE THERMAL CONDUCTIVITY, k(T). Start from the beginning of the section and read up to, but not including, Example 2-20. Focus on: The definition of k_avg in Equation 2-75. The specific result for k_avg when k(T) is a linear function (Equation 2-80). The conceptual point made in Figure 2-63: how a variable k leads to a non-linear temperature profile.

The Linear Case: A Vital Shortcut

As you just read, for the very common case where thermal conductivity varies linearly with temperature, $k (T) = k_{0} (1 + β T)$ , there's a fantastic shortcut. The average conductivity simplifies to:

$k_{a v g} = k_{0} (1 + β \frac{T _{1} + T _{2}}{2}) = k (T_{a v g})$

This means for linear dependencies, you don't even need to integrate! You can simply calculate the arithmetic average of the boundary temperatures, $T_{a v g} = (T_{1} + T_{2}) /2$ , find the conductivity at that average temperature, $k (T_{a v g})$ , and use this value in the standard heat flow equation.

Let's see this shortcut in a numerical problem.

HEAT CONDUCTION EQUATION

The same textbook provides a worked example that directly applies this shortcut.

Please review EXAMPLE 2–21 (Heat Conduction through a Wall with k(T)). Notice how they first calculate T_avg, then k_avg = k(T_avg), and finally use this value in the simple conduction formula to find the heat rate Q.

3. Handling Non-Linear Dependencies

The $k_{a v g} = k (T_{a v g})$ shortcut is only valid when $k (T)$ is a linear function. For any other dependency, such as $k (T) = α T^{2}$ or $k (T) = C / T$ , you must go back to the fundamental integral. This is a common setup for JEE Advanced questions.

Let's work through two examples that require direct integration.

Fourier Law Of Heat Conduction - Curio Physics

The webpage 'Fourier Law Of Heat Conduction' from Curio Physics has two excellent, challenging examples that are perfect for practice. They show exactly how to set up and solve the integrals for non-linear cases.

Please study Example 1 (where k = C/T) and Example 2 (where k = αT). Follow the integration step-by-step. For Example 2, focus only on the primary solution. We will discuss the 'Aliter' (alternative) solution next.

A Note on Common Pitfalls

Let's analyze the "Aliter" method shown in Example 2 of the Curio Physics article. It attempts to use an average temperature $T_{a v g}$ and an average temperature gradient $\frac{d T}{d x}$ . This approach is conceptually incorrect and will lead to the wrong answer for most non-linear dependencies.

Why is it wrong? It assumes $\frac{d T}{d x} = \frac{T _{2} - T _{1}}{l}$ . This is only true if the temperature profile $T (x)$ is a straight line, which happens only when $k$ is constant. But the entire premise of this problem is that $k$ varies with $T$ , which in turn varies with $x$ . As shown in Figure 2-63 from the textbook, this results in a curved temperature profile. The derivative of a curve is not constant.

Key takeaway for problem-solving: Unless $k (T)$ is linear, do not use average temperature or average gradient shortcuts. Always rely on the direct integration method:
$H = \frac{A}{L} \int_{T_{2}}^{T_{1}} k (T) d T$

Test your understanding!

A plane wall of thickness $L$ and area $A$ has its faces maintained at temperatures $T_{1}$ and $T_{2}$ ( $T_{1} > T_{2}$ ). The thermal conductivity of the material is given by $k (T) = α T^{2}$ , where $α$ is a constant. Find the rate of heat flow, $H$ , through the wall.

Show answer

We start from the general formula we derived:
$H = \frac{A}{L} \int_{T_{2}}^{T_{1}} k (T) d T$
Substitute $k (T) = α T^{2}$ :
$H = \frac{A}{L} \int_{T_{2}}^{T_{1}} α T^{2} d T$
Since $α$ is a constant, we can pull it out of the integral:
$H = \frac{α A}{L} \int_{T_{2}}^{T_{1}} T^{2} d T$
Now, we perform the integration:
$H = \frac{α A}{L} [\frac{T ^{3}}{3}]_{T_{2}}^{T_{1}}$
Evaluate at the limits:
$H = \frac{α A}{3 L} (T_{1}^{3} - T_{2}^{3})$
This is the final expression for the heat flow. Notice how the structure of the solution follows the same pattern as the examples.

4. Generalizing the Method: Position-Dependent Conductivity

The powerful integration technique we've been using is not limited to temperature dependence. It works just as well if the conductivity varies with position, $k (x)$ . While this is less common physically, it's a valid problem type to test your mathematical modeling skills.

In this case, Fourier's Law is $H = - k (x) A \frac{d T}{d x}$ .
Separating variables now keeps $k (x)$ on the left side with $d x$ :

$H \frac{d x}{k ( x )} = - A d T$

Integrating from $x = 0$ to $x = L$ and $T_{1}$ to $T_{2}$ :

$H \int_{0}^{L} \frac{d x}{k ( x )} = - A \int_{T_{1}}^{T_{2}} d T = A (T_{1} - T_{2})$

So, the heat current is:
$H = \frac{A ( T _{1} - T _{2} )}{\int _{0}^{L} \frac{d x}{k ( x )}}$
Notice the similarity to thermal resistance. The term in the denominator is effectively the total resistance of the slab.

The following video shows a complete worked example for a linear position-dependent conductivity, $k (x) = a + b x$ .

Physics 24 Heat Transfer: Conduction (17 of 34) Single Path - Variable Conductivity

This video, 'Single Path - Variable Conductivity' by Michel van Biezen, demonstrates how to solve a problem where thermal conductivity is a function of position, k(x). It's a great illustration of how the same 'separate and integrate' method works for a different type of variable dependency.

Watch the entire video. Observe how the presenter sets up Fourier's Law with k(x), separates the variables dx and dT, and integrates both sides with their respective limits to solve for the heat flow DQ/DT.

Conclusion

Today we've extended our analysis of heat conduction to the more realistic and challenging case of variable thermal conductivity. The key was to recognize that this variation turns the problem into a simple exercise in integral calculus.

Key Takeaways:

For any 1D steady-state problem with variable conductivity, the master strategy is to separate variables in Fourier's Law and integrate over the boundaries.
For a plane wall where $k$ depends on temperature, the heat current is $H = \frac{A}{L} \int_{T_{2}}^{T_{1}} k (T) d T$ .
If $k (T)$ is a linear function, you can use the powerful shortcut of calculating $k_{a v g} = k (T_{a v g})$ and using it in the standard conduction formula.
If $k (T)$ is non-linear, you must perform the integration directly. Be wary of incorrect shortcuts.
Variable conductivity causes the temperature profile across a material to be non-linear, even in a simple plane wall.

Preview of the Next Lesson:
We've spent a lot of time on steady-state conduction. But what happens when an object is initially hot and cools down over time? This is a transient problem. In our next lesson, we will begin our study of convection by looking at Newton's Law of Cooling, which governs this exact process. We will learn how to apply Newton's law of cooling to solve transient cooling and heating problems.

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