Question

Consider two infinitely long fins made of the same material and exposed to the same convective environment. One fin has a square cross-section of side \(a\), and the other has a circular cross-section of diameter \(d\). Assume \(a = d\). The ratio of the steady-state heat transfer rate from the square fin to that from the circular fin, \(\frac{\dot{Q}_{\text{square}}}{\dot{Q}_{\text{circular}}}\), is:

• A. \(\frac{\pi}{4}\)
• B. \(\frac{4}{\pi}\)
• C. \(\frac{2}{\sqrt{\pi}}\)
• D. \(\frac{\sqrt{\pi}}{2}\)

Hint:

For fin comparison problems, identify the constant parameters first. If only the shape changes, \(\dot{Q} \propto \sqrt{P A_c}\) for infinite fins, and \(\dot{Q} \propto \sqrt{P A_c} \tanh(mL)\) for finite fins with insulated tips.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to compare the heat transfer rates of two infinitely long fins: one with a square cross-section and one with a circular cross-section. The characteristic dimensions are equal (side \(a\) = diameter \(d\)). Both fins have the same thermal conductivity \(k\) and convective heat transfer coefficient \(h\). Step 2: Key Formula:
The rate of heat transfer from an infinitely long fin is given by: \[ \dot{Q} = \sqrt{hPkA_c} \, \theta_0 \] where:

• \(h\) = Convective heat transfer coefficient (constant for both)
• \(P\) = Perimeter of the cross-section
• \(k\) = Thermal conductivity (constant for both)
• \(A_c\) = Cross-sectional area
• \(\theta_0\) = Temperature difference at the base (constant for both)

Step 3: Calculation:
Since \(h, k, \text{ and } \theta_0\) are the same for both fins, the ratio depends only on the geometric parameters \(\sqrt{P A_c}\). \[ \frac{\dot{Q}_{\text{square}}}{\dot{Q}_{\text{circular}}} = \frac{\sqrt{P_s A_{cs}}}{\sqrt{P_c A_{cc}}} \] For the Square Fin (Side \(a\)): Perimeter \(P_s = 4a\)
Area \(A_{cs} = a^2\)
Geometric factor: \[ P_s A_{cs} = (4a)(a^2) = 4a^3 \] For the Circular Fin (Diameter \(d = a\)): Perimeter \(P_c = \pi d = \pi a\)
Area \(A_{cc} = \frac{\pi}{4} d^2 = \frac{\pi}{4} a^2\)
Geometric factor: \[ P_c A_{cc} = (\pi a) \left( \frac{\pi}{4} a^2 \right) = \frac{\pi^2}{4} a^3 \] Taking the Ratio: \[ \text{Ratio} = \sqrt{\frac{4a^3}{\frac{\pi^2}{4} a^3}} = \sqrt{\frac{16}{\pi^2}} = \frac{4}{\pi} \] Step 4: Final Answer:
The ratio is \(\frac{4}{\pi}\).