Let's break down and solve the problem step by step. --- ## **Given Data** - **Fin length (\(L\))**: 40 mm = 0.04 m - **Fin diameter (\(d\))**: 1.4 mm = 0.0014 m - **Fin material thermal conductivity (\(k_{Al}\))**: 140 W/m·K - **Base temperature (\(T_b\))**: 50°C - **Ambient temperature (\(T_\infty\))**: 25°C - **Heat transfer coefficient (\(h\))**: 1000 W/m²·K - **Adiabatic tip condition** --- ## **(a) Fin Heat Transfer Rate (q)** ### **Step 1: Calculate Fin Parameters** - **Perimeter, \(P\):** \[ P = \pi d = \pi \times 0.0014 = 0.004398 \text{ m} \] - **Cross-sectional area, \(A_c\):** \[ A_c = \frac{\pi d^2}{4} = \frac{\pi \times (0.0014)^2}{4} = 1.539 \times 10^{-6} \text{ m}^2 \] --- ### **Step 2: Calculate Fin Parameter \(m\)** \[ m = \sqrt{\frac{hP}{kA_c}} \] Plug in the numbers: \[ m = \sqrt{ \frac{1000 \times 0.004398}{140 \times 1.539 \times 10^{-6}} } \] \[ m = \sqrt{ \frac{4.398}{2.1546 \times 10^{-4}} } \] \[ m = \sqrt{20425} = 142.9 \text{ m}^{-1} \] --- ### **Step 3: Fin Heat Transfer Rate Equation (Adiabatic Tip)** \[ q_{fin} = \sqrt{hPkA_c} (T_b - T_\infty) \tanh(mL) \] Calculate \(\sqrt{hPkA_c}\): \[ \sqrt{hPkA_c} = \sqrt{1000 \times 0.004398 \times 140 \times 1.539 \times 10^{-6}} \] \[ = \sqrt{0.000951} = 0.03085 \] Now, \(mL = 142.9 \times 0.04 = 5.716\) \[ \tanh(5.716) \approx 0.99998 \approx 1 \] So, \[ q_{fin} = 0.03085 \times (50 - 25) \times 1 = 0.03085 \times 25 = 0.771 \text{ W} \] --- ### **Final Answer for (a):** **The fin heat transfer rate is approximately:** \[ \boxed{0.77 \text{ W}} \] --- ## **(b) Paint Coating Thickness for Maximum Heat Transfer Rate** ### **Step 1: Equivalent Conductivity for Composite Fin** Now, the fin is coated with paint of conductivity \(k_{paint} = 2.5\) W/m·K. The **overall thermal conductivity** of the coated fin will be much lower than aluminum. ### **Step 2: Critical Radius Concept** For a **cylinder**, the heat transfer is maximized when the outer radius \(r_o\) is: \[ r_{cr} = \frac{k_{paint}}{h} \] Given: - \(k_{paint} = 2.5\) W/m·K - \(h = 1000\) W/m²·K So, \[ r_{cr} = \frac{2.5}{1000} = 0.0025 \text{ m} = 2.5 \text{ mm} \] The original radius is: \[ r_{i} = \frac{d}{2} = 0.0007 \text{ m} = 0.7 \text{ mm} \] ### **Step 3: Coating Thickness** \[ \text{Thickness} = r_{cr} - r_{i} = 2.5 \text{ mm} - 0.7 \text{ mm} = 1.8 \text{ mm} \] --- ### **Step 4: Painted Fin Heat Transfer Rate** Now, the new outer diameter = \(2 \times 2.5 = 5.0\) mm - **Perimeter**: \(P = \pi d_{o} = \pi \times 0.005 = 0.0157 \) m - **Cross-sectional area of paint + aluminum**: Use the area at \(r_o\): \[ A_c = \pi (r_o^2 - r_i^2) = \pi ((0.0025)^2 - (0.0007)^2) = \pi (6.25 \times 10^{-6} - 0.49 \times 10^{-6}) = \pi (5.76 \times 10^{-6}) = 1.81 \times 10^{-5} \text{ m}^2 \] - **Thermal conductivity for the fin**: Use \(k_{paint} = 2.5\) W/m·K (since paint is outermost and dominant) - **Calculate new \(m\):** \[ m_{paint} = \sqrt{\frac{hP}{k_{paint}A_c}} \] \[ = \sqrt{\frac{1000 \times 0.0157}{2.5 \times 1.81 \times 10^{-5}}} \] \[ = \sqrt{\frac{15.7}{4.525 \times 10^{-5}}} = \sqrt{347,022} = 589 \text{ m}^{-1} \] - **Fin length \(L = 0.04\) m, so \(mL = 589 \times 0.04 = 23.56\), \(\tanh(23.56) \approx 1\)** - **Now, \(\sqrt{hPk_{paint}A_c}\):** \[ \sqrt{1000 \times 0.0157 \times 2.5 \times 1.81 \times 10^{-5}} = \sqrt{0.000710} = 0.0267 \] - **Painted fin heat transfer rate:** \[ q_{fin,paint} = 0.0267 \times 25 = 0.667 \text{ W} \] --- ### **Final Answers for (b):** - **Paint thickness for maximum heat transfer:** \[ \boxed{1.8 \text{ mm}} \] - **Maximum painted fin heat transfer rate:** \[ \boxed{0.67 \text{ W}} \] --- ## **Summary Table** | Part | Quantity | Value | |------|--------------------------------------------|--------------| | (a) | Fin heat transfer rate (aluminum) | 0.77 W | | (b) | Paint thickness for max heat transfer | 1.8 mm | | (b) | Painted fin heat transfer rate (max) | 0.67 W | --- Let me know if you'd like any step clarified or explained in detail!