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 rate before and after the replacement is

The correct option/answer is 4/π.

Step-by-Step Explanation:

For an infinitely long (very long) fin of uniform cross-section, the rate of heat transfer from the fin is given by the formula:
$$Q=\sqrt{h\cdot P\cdot k\cdot A_c}\cdot (T_b-T_{\infty })$$ where:
• $h$ is the heat transfer coefficient,
• $P$ is the perimeter of the fin,
• $k$ is the thermal conductivity of the fin material,
• $A_c$ is the cross-sectional area of the fin, and
• $(T_b-T_{\infty })$ is the temperature difference between the base of the fin and the ambient fluid.

Since the material, the heat transfer coefficient, and the temperature conditions are identical for both fins, the values of $h$, $k$, and $(T_b-T_{\infty })$ remain constant. Therefore, the heat transfer rate is directly proportional to the square root of the product of the perimeter and the cross-sectional area:
$$Q\propto \sqrt{P\cdot A_c}$$

Let the side length of the square fin be $a$, which is equal to the diameter $d$ of the circular fin (so, $d=a$).

1. Square Fin (Before Replacement):
• Perimeter, $P_1=4a$
• Cross-sectional area, $A_{c1}=a^2$

Thus, the product for the square fin is:
$$P_1{A}_{c1}=4a\cdot a^2=4a^3$$

2. Circular Fin (After Replacement):
• Diameter, $d=a$
• Perimeter, $P_2=\pi d=\pi a$
• Cross-sectional area, $A_{c2}=\frac{\pi }{4}{d}^2=\frac{\pi }{4}{a}^2$

Thus, the product for the circular fin is:
$$P_2{A}_{c2}=\left(\pi a\right)\cdot \left(\frac{\pi }{4}{a}^2\right)=\frac{{\pi }^2}{4}{a}^3$$

3. Ratio of Heat Transfer Rates:
The ratio of the heat transfer rate before replacement ($Q_1$) to after replacement ($Q_2$) is:
$$\frac{Q_1}{Q_2}=\sqrt{\frac{P_1{A}_{c1}}{P_2{A}_{c2}}}$$

Substitute the derived expressions into the ratio:
$$\frac{Q_1}{Q_2}=\sqrt{\frac{4a^3}{\frac{{\pi }^2}{4}{a}^3}}=\sqrt{\frac{16}{{\pi }^2}}=\frac{4}{\pi }$$

Thus, the ratio of heat transfer rate before and after replacement is indeed 4/π.