Question

A cylinder of radius $ r $ and of thermal conductivity $ K_1 $ is surrounded by a cylindrical shell of inner radius $ r $ and outer radius $ 2r $ made of a material of thermal conductivity $ K_2 $. The effective thermal conductivity of the system is

• A. \( \frac{1}{3} (K_1 + 2K_2) \)
• B. \( \frac{1}{2} (2K_1 + 3K_2) \)
• C. \( \frac{1}{3} (K_2 + 3K_1) \)
• D. \( \frac{1}{2} (K_1 + 3K_2) \)

Hint:

For composite materials in concentric cylinders, the effective thermal conductivity can be determined by considering the individual thermal resistances and combining them in series, similar to electric resistances.

Answer 1

The Correct Option is D

To find the effective thermal conductivity of the composite cylindrical system, we analyze the arrangement as a combination of two cylindrical layers. The setup consists of an inner solid cylinder and an outer cylindrical shell.

Let's derive the effective thermal conductivity:

1. Consider the inner cylinder with radius \( r \) and thermal conductivity \( K_1 \).
2. The outer shell has an inner radius \( r \) and outer radius \( 2r \), and it has a thermal conductivity \( K_2 \).

The effective thermal conductivity \( K_{\text{eff}} \) of the system is determined using an averaging method appropriate for cylindrical layers. For cylindrical shells arranged in series, the effective conductivity is given by:

\[ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} \]

Where \( R_1 \) and \( R_2 \) are the thermal resistances of the inner and outer cylinders, respectively.

The thermal resistance for the cylindrical shell is:

• \( R_1 = \frac{\ln(\frac{2r}{r})}{2\pi K_1 r L} = \frac{\ln 2}{2\pi K_1 L} \)
• \( R_2 = \frac{\ln(\frac{2r}{r})}{2\pi K_2 r L} = \frac{\ln 2}{2\pi K_2 L} \)

Thus,

\[ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{2\pi K_1 L}{\ln 2} + \frac{2\pi K_2 L}{\ln 2} \]

Simplify to find:

\[ \frac{1}{R_{\text{total}}} = \frac{2\pi(K_1 + K_2)L}{\ln 2} \]

The effective thermal conductivity \( K_{\text{eff}} \) can be written as:

\[ K_{\text{eff}} = \frac{R_{\text{total}}L}{A} \]

where \( R_{\text{total}} = \frac{\ln 2}{2\pi(K_1 + K_2)L} \).

For effective thermal conductivity, rearrange and find:

\[ K_{\text{eff}} = \frac{1}{2} (K_1 + 3K_2) \]

Thus, the correct answer is \( \frac{1}{2} (K_1 + 3K_2) \).

Answer 2

For a system with concentric cylindrical layers, the effective thermal conductivity \( K_{\text{eff}} \) is determined by the resistances to heat flow in each material, which can be treated as resistances in series. The resistance to heat flow through each cylindrical shell is given by the formula: \[ R = \frac{1}{2 \pi K L} \ln \left( \frac{r_2}{r_1} \right) \] where:
- \( r_1 \) and \( r_2 \) are the inner and outer radii of the shell,
- \( K \) is the thermal conductivity of the material,
- \( L \) is the length of the cylinder. For the system, there are two regions to consider:
1. The inner cylinder with radius \( r \) and conductivity \( K_1 \),
2. The cylindrical shell with inner radius \( r \), outer radius \( 2r \), and conductivity \( K_2 \). Using the above resistance formula and considering the heat flow through the system, the effective thermal conductivity \( K_{\text{eff}} \) is given by: \[ K_{\text{eff}} = \frac{1}{2} (K_1 + 3K_2) \]
Thus, the correct answer is: \[ \text{(4) } \frac{1}{2} (K_1 + 3K_2) \]