Heat Conduction in Variable Area Systems

Hello! Welcome to our next lesson on heat conduction.

In our previous session, we developed a powerful toolkit for analyzing complex arrangements of conductors using the thermal resistance analogy. We treated discrete conducting rods like resistors in a circuit, applying principles like Kirchhoff's laws to solve for junction temperatures and heat flows. This approach, however, relies on the assumption that each conductor has a uniform cross-sectional area, leading to the simple formula $R_{t h} = L / k A$ .

Today, we will address this lesson's learning outcome: to apply heat conduction principles to problems with variable cross-sectional areas. We'll move beyond the algebraic methods of combining discrete resistances and employ calculus to handle continuous variations in geometry. This is a crucial step towards modeling more realistic physical systems and a type of problem frequently featured in JEE Advanced. Your strong foundation in calculus will be a significant asset here.

1. The Fundamental Principle: Fourier's Law in Differential Form

When the cross-sectional area $A$ of a conductor is not constant, the temperature gradient $\frac{d T}{d x}$ will also vary along the length of the conductor, even in a steady state. Therefore, we cannot use $Δ T / L$ and must return to the more fundamental, differential form of Fourier's Law:

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

Here, $H$ is the heat current (rate of heat flow), $k$ is the thermal conductivity, $A (x)$ is the cross-sectional area as a function of position $x$ , and $\frac{d T}{d x}$ is the temperature gradient at that position.

The most critical concept for steady-state conduction is that the heat current $H$ must be constant at every point along the conductor. If it weren't, heat would be accumulating in or depleting from some part of the conductor, which contradicts the definition of a steady state.

This leads to a powerful insight. Since $H$ and $k$ (assuming it's constant) are fixed, we have:

$A (x) \frac{d T}{d x} = Constant$

This means that as the cross-sectional area $A (x)$ increases, the magnitude of the temperature gradient $\frac{d T}{d x}$ must decrease proportionally. The temperature will drop more slowly in wider sections and more steeply in narrower sections. The resulting temperature profile $T (x)$ will be a curve, not a straight line.

Analysis of an Axisymmetric Shape

The document from Clarkson University has a concise explanation and a helpful schematic of this exact point. It's a great way to build the core intuition before we dive into calculations.

Please read the analysis for PROBLEM 2.1. Focus on the logic that leads to the conclusion A_x * dT/dx = Constant and how this determines the shape of the temperature distribution curve shown in the schematic.

2. Case Study 1: Linearly Varying Area (The Truncated Cone)

A classic example of a variable cross-section is a truncated cone. Let's consider a rod whose radius changes linearly from $r_{1}$ to $r_{2}$ over a length $L$ .

To find the heat current $H$ , we follow a standard procedure that leverages your calculus skills:

Express Area as a Function of Position: Define the radius $r (x)$ as a linear function of $x$ . The area is then $A (x) = π [r (x)]^{2}$ .
Apply Fourier's Law: Substitute $A (x)$ into the differential equation $H = - k A (x) \frac{d T}{d x}$ .
Separate Variables: Rearrange the equation to get all terms involving $x$ on one side (with $d x$ ) and all terms involving $T$ on the other (with $d T$ ).
Integrate: Integrate both sides over the appropriate limits: from $x = 0$ to $x = L$ on one side, and from $T = T_{1}$ to $T = T_{2}$ on the other.
Solve for H: The result of the integration will be an equation relating $H$ to the known geometric and thermal properties.

This exact process is demonstrated very clearly in the following video.

Physics 24 Heat Transfer: Conduction (16 of 34) Single Path - Variable Cross-Section

This video by Michel van Biezen works through a complete example of a conductor with a linearly changing cross-sectional area. It's a perfect illustration of the integration method.

Watch the entire video (00:00 - 07:52). Pay close attention to these key steps: Setting up the area as a linear function of x (00:40). Writing Fourier's Law in differential form, noting H (dQ/dt) is constant (01:50). Separating the variables (03:10) and integrating to find the final answer.

Deriving the Equivalent Thermal Resistance

After integrating, we can rearrange the final expression for $H$ into the familiar form $H = \frac{Δ T}{R _{e q}}$ . For a truncated cone with end radii $r_{1}$ and $r_{2}$ and length $L$ , the integration yields a heat current of:

$H = \frac{k π r _{1} r _{2}}{L} (T_{1} - T_{2})$

From this, we can directly identify the equivalent thermal resistance as:

$R_{e q} = \frac{L}{k π r _{1} r _{2}}$

This is a very useful result to remember for JEE problems, as it can save you the time of re-deriving it from scratch. Notice that if $r_{1} = r_{2} = r$ , this formula simplifies to $R = \frac{L}{k π r ^{2}} = \frac{L}{k A}$ , the standard formula for a uniform cylinder.

3. Case Study 2: Radially Varying Area (Cylindrical and Spherical Shells)

Another common geometry is heat flow in the radial direction, for instance, through the wall of a pipe (a hollow cylinder) or a spherical shell.

Hollow Cylinder

Consider a hollow cylinder of length $L$ , with inner radius $r_{1}$ (at temperature $T_{1}$ ) and outer radius $r_{2}$ (at temperature $T_{2}$ ). Heat flows radially outward.

The key difference is that the variable is now the radius, $r$ . The area through which heat flows is the surface area of a cylinder at radius $r$ , which is:
$A (r) = 2 π r L$
Fourier's law is now written in terms of $r$ :
$H = - k A (r) \frac{d T}{d r} = - k (2 π r L) \frac{d T}{d r}$
The method is the same: separate variables and integrate.

Physics 24.1 Variable Heat Transfer (20 of 25) Cylindrical Wall Conductivity

Let's watch another video from the same instructor that derives the heat flow through a cylindrical wall.

Watch from the beginning to 05:33. The logic is identical to the previous case, but applied to a different geometry. Focus on how the area A(r) = 2πRL is defined and how the integration of dr/r leads to a natural logarithm in the final expression.

The result of the integration gives the equivalent thermal resistance for a hollow cylinder:

$R_{e q} = \frac{l n ( r _{2} / r _{1} )}{2 π k L}$

This is another essential formula for your problem-solving arsenal.

Test your understanding!

Now consider a hollow spherical shell with inner radius $r_{1}$ and outer radius $r_{2}$ . Heat flows radially outward from the inner surface to the outer surface.

What is the correct expression for the area $A (r)$ through which heat flows at a radius $r$ ?
Set up the integral to find the thermal resistance. You don't need to solve it, just write down the definite integral.

Show answer

For a sphere, the surface area at a radius $r$ is $A (r) = 4 π r^{2}$ .
We start with Fourier's Law: $H = - k A (r) \frac{d T}{d r}$ .
Separating variables gives: $H \frac{d r}{A ( r )} = - k d T$ .
Substituting for $A (r)$ and rearranging for resistance: $\frac{d T}{H} = - \frac{d r}{k ( 4 π r ^{2} )}$ .
The resistance $R$ is $\frac{Δ T}{H}$ . To find it, we integrate the right side:
$R_{e q} = \int_{r_{1}}^{r_{2}} \frac{d r}{k ( 4 π r ^{2} )} = \frac{1}{4 π k} \int_{r_{1}}^{r_{2}} \frac{d r}{r ^{2}}$
Solving this integral gives $R_{e q} = \frac{1}{4 π k} (\frac{1}{r _{1}} - \frac{1}{r _{2}})$ .

4. Problem-Solving Strategy for Complex Geometries

JEE Advanced problems often combine concepts. You might encounter a system composed of several parts, some with uniform areas and others with variable areas. The strategy is to break the system down and use your knowledge of thermal circuits.

Example Problem:
A composite rod is made of two sections in series.

Section 1: A solid cylinder of length $L_{1}$ , radius $r$ , and conductivity $k_{1}$ .
Section 2: A truncated cone of length $L_{2}$ , with end radii $r$ and $2 r$ , and conductivity $k_{2}$ .

How to find the total heat current $H$ for a temperature difference $Δ T$ across the entire composite rod?

Model as a Circuit: This is a series combination. The total resistance is $R_{t o t a l} = R_{1} + R_{2}$ .
Calculate $R_{1}$ : This is a uniform cylinder. Use the standard formula:
$R_{1} = \frac{L _{1}}{k _{1} A} = \frac{L _{1}}{k _{1} π r ^{2}}$
Calculate $R_{2}$ : This is a truncated cone. Use the formula we derived today:
$R_{2} = \frac{L _{2}}{k _{2} π r _{1} r _{2}} = \frac{L _{2}}{k _{2} π ( r ) ( 2 r )} = \frac{L _{2}}{2 k _{2} π r ^{2}}$
Find Total Resistance: Add them up: $R_{t o t a l} = \frac{L _{1}}{k _{1} π r ^{2}} + \frac{L _{2}}{2 k _{2} π r ^{2}}$ .
Calculate Heat Current: $H = \frac{Δ T}{R _{t o t a l}}$ .

This modular approach—calculating the resistance of each part using the appropriate formula and then combining them using circuit rules—is the key to solving complex, multi-part problems.

Conclusion

Today we've extended our study of heat conduction to handle non-uniform geometries, a scenario that demands the application of calculus. This approach is fundamental to many areas of physics and engineering.

Key Takeaways:

For a variable cross-section $A (x)$ , you must use the differential form of Fourier's Law: $H = - k A (x) \frac{d T}{d x}$ .
In steady state, the heat current $H$ is constant at every point along the conductor. This implies that $A (x) \frac{d T}{d x}$ is also constant.
The general method is to express $A$ as a function of position, separate variables in Fourier's law, and integrate over the object's dimensions.
The equivalent thermal resistances for a truncated cone and for hollow cylindrical/spherical shells are standard results worth remembering to speed up problem-solving.
- Truncated Cone: $R_{e q} = \frac{L}{k π r _{1} r _{2}}$
- Hollow Cylinder: $R_{e q} = \frac{l n ( r _{2} / r _{1} )}{2 π k L}$
- Hollow Sphere: $R_{e q} = \frac{1}{4 π k} (\frac{1}{r _{1}} - \frac{1}{r _{2}})$

Preview of the Next Lesson:
We have now covered heat conduction in simple (linear) and complex (radial) geometries. In the next lesson, we will solidify these ideas by formally tackling steady-state heat conduction in cylindrical and spherical geometries. We will practice more problems involving these radial systems and ensure you are comfortable applying these important formulas in various contexts.

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