Question

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

• A. \(4/\pi\)
• B. \(16/\pi^2\)
• C. \(1/\pi\)
• D. \(1/\pi^2\)

Hint:

Write the long fin heat rate formula \(Q=\sqrt{hPkA}\,\theta_b\) and compare \(P\) and \(A\) for the two shapes.

Answer 1

The Correct Option is A

Step 1: Recall the heat rate formula for a very long fin.
For a fin long enough that its tip loses no heat, the heat rate is \(Q = \sqrt{hPkA}\,\theta_b\), where \(P\) is the cross-section perimeter, \(A\) is the cross-section area, \(k\) is the conductivity, \(h\) the heat transfer coefficient, and \(\theta_b\) the base excess temperature. Since \(h\), \(k\), and \(\theta_b\) stay the same before and after, the ratio of heat rates reduces to \(Q_{square}/Q_{circle} = \sqrt{(P_sA_s)/(P_cA_c)}\).

Step 2: Work out P and A for the square fin.
Let the side of the square be \(a\). Its perimeter is \(P_s = 4a\) and its area is \(A_s = a^2\), so \(P_sA_s = 4a^3\).

Step 3: Work out P and A for the circular fin.
The diameter equals the square's side, so \(d = a\). Its perimeter is \(P_c = \pi a\) and its area is \(A_c = \pi a^2/4\), so \(P_cA_c = \pi^2a^3/4\).

Step 4: Take the ratio.
\[ \frac{Q_{square}}{Q_{circle}} = \sqrt{\frac{4a^3}{\pi^2a^3/4}} = \sqrt{\frac{16}{\pi^2}} = \frac{4}{\pi} \]
Option (B) is what you get if you forget to take the square root, and (C) and (D) come from flipping the ratio the wrong way round.

Final Answer:
The square fin transfers more heat because it packs a larger perimeter-to-area product than a circle of the same characteristic length. \[ \boxed{\dfrac{4}{\pi}} \]