Here’s a breakdown of the problem based on the image: **Given:** - Long stainless steel rod - Thermal conductivity, k = 16 W/m·°C - Cross section: Equilateral triangle, side length = 15 mm - Length, L = 90 mm - Attached to wall at T_base = 450°C - Surroundings at T_∞ = 50°C - Surface conductance, h = 50 W/m²·°C **To Find:** Heat dissipated by the rod (Q). --- ### Step 1: Calculate Cross-sectional Area (A) For an equilateral triangle of side \( a = 15 \) mm = 0.015 m: \[ A = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \times (0.015)^2 = 9.74 \times 10^{-5} \text{ m}^2 \] --- ### Step 2: Perimeter (P) \[ P = 3 \times a = 3 \times 0.015 = 0.045 \text{ m} \] --- ### Step 3: Fin Parameter (m) \[ m = \sqrt{\frac{h P}{k A}} \] \[ m = \sqrt{\frac{50 \times 0.045}{16 \times 9.74 \times 10^{-5}}} \] \[ = \sqrt{\frac{2.25}{0.0015584}} \] \[ = \sqrt{1444.7} \] \[ \approx 38.01 \text{ m}^{-1} \] --- ### Step 4: Find mL \[ L = 90 \text{ mm} = 0.09 \text{ m} \] \[ mL = 38.01 \times 0.09 \approx 3.42 \] --- ### Step 5: Fin Heat Dissipation (Q) For a long fin (since mL > 2): \[ Q = \sqrt{h P k A} (T_{base} - T_\infty) \] \[ \sqrt{h P k A} = \sqrt{50 \times 0.045 \times 16 \times 9.74 \times 10^{-5}} \] \[ = \sqrt{0.0035088} \] \[ = 0.0593 \] \[ Q = 0.0593 \times (450 - 50) \] \[ = 0.0593 \times 400 \] \[ = 23.72 \text{ W} \] --- ### Step 6: Choose the Closest Option Given options: - 6.12 W - 312 kW - 1.7 W - 5.12 W 23.72 W is not among the options, but let’s check the calculation for errors. For a long fin, the equation is: \[ Q = \sqrt{h P k A} (T_{base} - T_\infty) \tanh(mL) \] Calculate \(\tanh(3.42)\): \[ \tanh(3.42) \approx 0.997 \] \[ Q = 0.0593 \times 400 \times 0.997 = 23.65 \text{ W} \] Again, much higher than the options, which suggests either an error in the question, unit mismatch, or a miscalculation in one of the steps. **Check the area and perimeter calculations again:** - \(A = \frac{\sqrt{3}}{4} a^2\) - \(a = 0.015\) m - \((0.015)^2 = 0.000225\) - \(A = 0.433 \times 0.000225 = 9.74 \times 10^{-5}\) m² (correct) - \(P = 3 \times 0.015 = 0.045\) m (correct) Everything checks out. ### **Final Thought** Assuming the provided data and equations are correct, the calculated heat dissipation is about **23.7 W**, which does not match any of the options. Please double-check the units and problem statement for any misinterpretation. If you must choose an answer, **6.12 W** is the closest, but it's significantly lower than the calculation. Let me know if you need further clarification or if any data is missing.