Question

Conductive thermal resistance (R) offered by a wall of thickness "L", area "A" and thermal conductivity "k" is given by which of following equations?

• A. R = LAk
• B. R = L/ kA
• C. R = k LA
• D. R = A Lk

Hint:

Thermal resistance $R = \frac{L}{kA}$ is directly proportional to thickness ($L$) and inversely proportional to thermal conductivity ($k$) and area ($A$).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Conduction is the transfer of heat through solid materials via molecular interaction.
Fourier's Law of heat conduction describes the rate of heat transfer through a solid barrier.
This relationship can be represented in a form analogous to Ohm's Law for electrical circuits, where heat flow represents current, temperature difference represents voltage, and thermal resistance represents electrical resistance.
Key Formula or Approach:
Fourier's Law of heat conduction through a flat wall is expressed as:
\[ Q = \frac{kA \Delta T}{L} \]
Where:
$Q$ is the rate of heat transfer,
$k$ is the thermal conductivity,
$A$ is the cross-sectional area,
$\Delta T$ is the temperature difference, and
$L$ is the thickness of the wall.

Step 2: Detailed Explanation:
We can rearrange Fourier's equation to isolate the temperature driving force ($\Delta T$):
\[ Q = \frac{\Delta T}{\left(\frac{L}{kA}\right)} \]
Comparing this to Ohm's Law ($I = \frac{V}{R}$):
The term in the denominator represents the resistance to heat transfer, known as conductive thermal resistance ($R$):
\[ R = \frac{L}{kA} \]
This equation indicates that thermal resistance increases with wall thickness ($L$) and decreases with higher thermal conductivity ($k$) or surface area ($A$).

Step 3: Final Answer
The correct option is (B).