Discussion
A very long fin of a uniform square cross-section is replaced by another very long fin of a uniform circular cross-section of the same material. Assume uniform and identical heat transfer coefficient for both the fins. If the diameter of the circular fin is equal to the side length of the square fin, then the ratio of heat transfer rates before and after the replacement is
Answer: A
The ratio is \(4/\pi\).
For a very long fin, \(q = \sqrt{hPkA_{c}}\,\theta_{b}\), so with the same material, coefficient and base temperature, \(q \propto \sqrt{PA_{c}}\). With side \(a\) and diameter \(a\):
\[ \begin{aligned} \text{square} &: P A_{c} = 4a \cdot a^{2} = 4a^{3} \\ \text{circle} &: P A_{c} = \pi a\cdot\frac{\pi a^{2}}{4} = \frac{\pi^{2}a^{3}}{4} \\ \frac{q_{sq}}{q_{circ}} &= \sqrt{\frac{16}{\pi^{2}}} = \frac{4}{\pi} \end{aligned} \]
\(16/\pi^{2}\) forgets the square root. \(1/\pi\) and \(1/\pi^{2}\) are below 1, but the square fin has more perimeter and more area than the inscribed circle, so it must transfer more heat.
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