Problem 29 - Master Resource Book in JEE Main Physics

Question

A cylinder of radius \(r\) and thermal conductivity \(K_{1}\) is surrounded by a cylindrical shell of linear radius \(r\) and outer radius \(2 r\), whose thermal conductivity is \(K_{2}\). There is no loss of heat across cylindrical surfaces, when the ends of the combined system are maintained at temperatures \(T_{1}\) and \(T_{2}\). The effective thermal conductivity of the system, in the steady state is (a) \(\frac{K_{1} K_{1}}{K_{1}+K_{2}}\) (b) \(K_{1}+K_{2}\) (c) \(\frac{K_{1}+3 K_{2}}{4}\) (d) \(\frac{3 K_{1}+K_{2}}{4}\)

Step-by-Step Solution

Answer

The effective thermal conductivity is \( \frac{K_1 + 3K_2}{4} \), option (c).

Step 1: Identify the parallel thermal resistances

The system consists of a solid cylinder with thermal conductivity \( K_{1} \) and a surrounding cylindrical shell with thermal conductivity \( K_{2} \). Both parts are subjected to the same temperature difference across their lengths. Since the heat flows parallel through each part, these contribute to a parallel thermal resistance system.

Step 2: Determine thermal resistance for each part

For the inner cylinder, the thermal resistance is given by \( R_{1} = \frac{L}{K_{1} imes ext{cross-sectional area}} = \frac{L}{K_{1} imes \pi r^2} \). For the outer shell, the thermal resistance is \( R_{2} = \frac{L}{K_{2} imes ext{cross-sectional area}} = \frac{1}{K_{2} imes \pi ( (2r)^2-r^2) } = \frac{L}{K_{2} imes 3\pi r^2} \).

Step 3: Apply formula for effective thermal conductivity for parallel resistances

The effective thermal conductivity \( K_{ ext{eff}} \) in a parallel arrangement is calculated using: \( \frac{1}{R_{ ext{eff}}} = \frac{1}{R_1} + \frac{1}{R_2} \). Substitute the expressions for \( R_1 \) and \( R_2 \) to get:\[\frac{K_{ ext{eff}} \times \pi(2r)^2}{L} = \frac{K_1 \times \pi r^2}{L} + \frac{K_2 \times \pi \times 3r^2}{L}\]

Step 4: Simplify the equation

Solving for \( K_{ ext{eff}} \):\[K_{ ext{eff}} \times 4 = K_1 + 3K_2 \K_{ ext{eff}} = \frac{K_1 + 3K_2}{4}\]This matches with the choice \( (c) \).

Key Concepts

Thermal resistance

Thermal resistance is a crucial concept in understanding heat transfer, much like electrical resistance in electrical circuits. It describes a material's ability to resist the flow of heat through it. A high thermal resistance means the material is good at insulating against heat flow, while a low thermal resistance indicates it allows heat to pass through more easily. In formulas, thermal resistance is often denoted by \( R \) and is calculated as the ratio of the temperature difference across the material to the heat flow through it. For a simple homogeneous material, it's given by \( R = \frac{L}{K \cdot A} \), where:

• \( L \) is the length through which heat is transferred.
• \( K \) is the material's thermal conductivity.
• \( A \) is the cross-sectional area.

Understanding thermal resistance helps in designing systems where controlling temperature is crucial, such as in building insulation, electronic components, and mechanical systems.

Parallel resistances

When dealing with thermal systems, understanding the concept of parallel resistances is essential for determining effective thermal conductivities. Just like in electrical circuits where resistors can be placed in parallel, in thermal systems, different paths for heat flow can be arranged parallel to each other.
In the original exercise, the inner cylinder and its surrounding shell act like parallel resistors for heat flow. The formula to find the combined effect of these paths involves adding the reciprocals of their resistances:

• \( \frac{1}{R_{\text{eff}}} = \frac{1}{R_1} + \frac{1}{R_2} \)

This equation is crucial because it helps determine how heat will flow through composite materials comprising different layers or paths. When dealing with such configurations, solutions need to consider how heat can take different routes simultaneously. Thus, it's important to account for these parallel resistances to calculate the overall heat transfer effectively.

Steady state heat transfer

Steady state heat transfer is a fundamental principle in thermal analysis. In this condition, the temperature distribution does not change with time, meaning any point within the material maintains a constant temperature after reaching equilibrium. In practical terms, once the system reaches steady state, the heat entering a section equals the heat leaving it, ensuring no accumulation.
This principle simplifies the analysis of heat transfer in systems. For example, in the exercise, the effective thermal conductivity could be computed without worrying about changes over time because it's assumed that the system remains consistent in temperature difference and thermal conditions.
Understanding steady state helps greatly in optimizing thermal systems, for example, in ensuring electronic devices do not overheat and maintaining buildings' internal climates efficiently. It assures that once equilibrium is reached, the system operates under predictable conditions, making it easier to manage and plan for sustainable heat transfer solutions.