Question

An infinitely long pin fin, attached to an isothermal hot surface, transfers heat at a steady rate of \( \dot{Q}_1 \) to the ambient air. If the thermal conductivity of the fin material is doubled, while keeping everything else constant, the rate of steady-state heat transfer from the fin becomes \( \dot{Q}_2 \). The ratio \( \dot{Q}_2 / \dot{Q}_1 \) is

• A. \( \sqrt{2} \)
• B. 2
• C. \( \frac{1}{\sqrt{2}} \)
• D. \( \frac{1}{2} \)

Hint:

In heat transfer problems involving pin fins, the rate of heat transfer is proportional to the square root of the thermal conductivity.

Answer 1

The Correct Option is A

For an infinitely long pin fin, the rate of heat transfer \( \dot{Q} \) is given by the following relationship: \[ \dot{Q} \propto \sqrt{k} \] where \( k \) is the thermal conductivity of the fin material. This equation indicates that the rate of heat transfer is proportional to the square root of the thermal conductivity.
If the thermal conductivity \( k \) of the fin material is doubled, i.e., \( k_2 = 2k_1 \), the new heat transfer rate \( \dot{Q}_2 \) becomes: \[ \dot{Q}_2 \propto \sqrt{2k_1} = \sqrt{2} \cdot \sqrt{k_1} \] Thus, the ratio of the heat transfer rates is: \[ \frac{\dot{Q}_2}{\dot{Q}_1} = \frac{\sqrt{2} \cdot \sqrt{k_1}}{\sqrt{k_1}} = \sqrt{2} \] Therefore, the ratio \( \dot{Q}_2 / \dot{Q}_1 \) is \( \sqrt{2} \). Final Answer: (A)