Fin Heat Transfer Rate Ratio Analysis

The heat transfer rate ($Q$) from a very long fin is given by the formula:

$Q = \sqrt{h P k A_c} (\theta_b)$

Where:

• $h$ = convective heat transfer coefficient
• $P$ = fin perimeter
• $k$ = thermal conductivity of the fin material
• $A_c$ = fin cross-sectional area
• $\theta_b$ = temperature difference between the fin base and the surrounding fluid

Since $h$, $k$, and $\theta_b$ are identical for both fins, the ratio of heat transfer rates depends only on the term $\sqrt{P A_c}$.

Calculating Geometric Properties

Let $s$ be the side length of the square fin and $d$ be the diameter of the circular fin. We are given that $d = s$.

Square Fin (Before Replacement)

• Cross-sectional Area, $A_{c, \text{square}}$ = $s \times s = s^2$
• Perimeter, $P_{\text{square}}$ = $4s$
• The term $\sqrt{P A_c}$ for the square fin is: $ \sqrt{P_{\text{square}} A_{c, \text{square}}} = \sqrt{(4s)(s^2)} = \sqrt{4s^3} $

Circular Fin (After Replacement)

• Diameter, $d$ = $s$
• Radius, $r$ = $d/2 = s/2$
• Cross-sectional Area, $A_{c, \text{circular}}$ = $\pi r^2 = \pi (s/2)^2 = \frac{\pi s^2}{4}$
• Perimeter, $P_{\text{circular}}$ = $\pi d = \pi s$
• The term $\sqrt{P A_c}$ for the circular fin is: $ \sqrt{P_{\text{circular}} A_{c, \text{circular}}} = \sqrt{(\pi s)\left(\frac{\pi s^2}{4}\right)} = \sqrt{\frac{\pi^2 s^3}{4}} $

Determining the Heat Transfer Ratio

The ratio of heat transfer rates before (square) and after (circular) is:

$ \frac{Q_{\text{before}}}{Q_{\text{after}}} = \frac{Q_{\text{square}}}{Q_{\text{circular}}} = \frac{\sqrt{h P_{\text{square}} k A_{c, \text{square}}} (\theta_b)}{\sqrt{h P_{\text{circular}} k A_{c, \text{circular}}} (\theta_b)} $

Simplifying the ratio using the $\sqrt{P A_c}$ terms:

$ \frac{Q_{\text{square}}}{Q_{\text{circular}}} = \sqrt{\frac{P_{\text{square}} A_{c, \text{square}}}{P_{\text{circular}} A_{c, \text{circular}}}} = \sqrt{\frac{4s^3}{\pi^2 s^3 / 4}} $

$ \frac{Q_{\text{square}}}{Q_{\text{circular}}} = \sqrt{\frac{4 \times 4}{\pi^2}} = \sqrt{\frac{16}{\pi^2}} = \frac{4}{\pi} $

Therefore, the ratio of heat transfer rates before and after the replacement is $4/\pi$.