Let's break down the problem step by step: --- ## **Given Data** - Rod length, \( L = 20 \) cm = **0.2 m** - Thermal conductivity, \( k = 50 \) W/m·K - Wall temperature, \( T_b = 200^\circ \)C - Ambient fluid temperature, \( T_\infty = 40^\circ \)C - Rod diameter, \( D = 5 \) cm = **0.05 m** - Heat transfer coefficient, \( h = 25 \) W/m²·K --- ## **Step 1: Area and Perimeter of the Rod** - **Perimeter, \( P \):** \[ P = \pi D = \pi \times 0.05 = 0.1571 \text{ m} \] - **Cross-sectional Area, \( A_c \):** \[ A_c = \frac{\pi D^2}{4} = \frac{\pi (0.05)^2}{4} = 0.00196 \text{ m}^2 \] --- ## **Step 2: Fin Parameter, \( m \)** \[ m = \sqrt{\frac{hP}{kA_c}} \] Plug in the values: \[ m = \sqrt{ \frac{25 \times 0.1571}{50 \times 0.00196} } \] Calculate numerator and denominator: - Numerator: \( 25 \times 0.1571 = 3.9275 \) - Denominator: \( 50 \times 0.00196 = 0.098 \) So: \[ m = \sqrt{ \frac{3.9275}{0.098} } = \sqrt{40.14} = 6.34 \text{ m}^{-1} \] --- ## **Step 3: Heat Transfer from the Fin (Rod)** Since the rod is **long** and exposed at the tip, use the following formula for a fin with an **adiabatic tip**: \[ q_{\text{fin}} = \sqrt{hPkA_c}(T_b - T_\infty) \tanh(mL) \] First, calculate \(\sqrt{hPkA_c}\): - \( h P k A_c = 25 \times 0.1571 \times 50 \times 0.00196 \) - \( = 3.9275 \times 50 \times 0.00196 \) - \( = 196.375 \times 0.00196 = 0.3849 \), then \( \sqrt{0.3849} = 0.6204 \) Now, calculate \( mL \): \[ mL = 6.34 \times 0.2 = 1.268 \] \[ \tanh(1.268) \approx 0.852 \] Plug in all values: \[ q_{\text{fin}} = 0.6204 \times (200 - 40) \times 0.852 \] \[ = 0.6204 \times 160 \times 0.852 \] \[ = 0.6204 \times 136.32 \] \[ = 84.6 \text{ W} \] --- ### **Final Answer for Part (a):** \[ \boxed{q_{\text{fin}} = 84.6 \text{ W}} \] --- ## **Step 4: Fin Efficiency (\( \eta_{\text{fin}} \))** Fin efficiency: \[ \eta_{\text{fin}} = \frac{q_{\text{fin}}}{h P L (T_b - T_\infty)} \] First, calculate denominator: \[ h P L (T_b - T_\infty) = 25 \times 0.1571 \times 0.2 \times 160 \] - \( 25 \times 0.1571 = 3.9275 \) - \( 3.9275 \times 0.2 = 0.7855 \) - \( 0.7855 \times 160 = 125.68 \) So, \[ \eta_{\text{fin}} = \frac{84.6}{125.68} = 0.673 \text{ or } 67.3\% \] --- ### **Final Answer for Part (b):** \[ \boxed{\eta_{\text{fin}} = 67.3\%} \] --- ## **Summary Table** | Part | Formula Used | Answer | |-------------|--------------|--------| | (a) Heat transfer dissipated, \(q_{\text{fin}}\) | \( q_{\text{fin}} = \sqrt{hPkA_c}(T_b - T_\infty) \tanh(mL) \) | \( \boxed{84.6 \text{ W}} \) | | (b) Fin efficiency, \( \eta_{\text{fin}} \) | \( \eta_{\text{fin}} = \frac{q_{\text{fin}}}{h P L (T_b - T_\infty)} \) | \( \boxed{67.3\%} \) | --- If you need clarification on any of the steps or formulas, let me know!