# Step-by-Step Solution: Handle as a Fin Let's break down the problem and solve it step-by-step: --- ## **1. Problem Restatement** - **Handle:** 16 in. long steel rod, 1/4 in. diameter, bent into a "U" and welded to a furnace door. - **Door temperature:** 160°C - **Room air temperature:** 24°C - **Convective cooling coefficient (h):** 12 W/m²·°C - **Thermal conductivity of steel (k):** 52 W/m·°C - **Assume:** 65°C or hotter will cause burns. - **Model:** Treat as two 8-inch (.2032 m) insulated-tip fins due to symmetry. --- ## **2. Fin Equation Setup** ### **Fin Parameters** - **Length of each fin (L):** 8 in = .2032 m - **Diameter (d):** .25 in = .00635 m - **Perimeter (P):** \( P = \pi d = \pi \times .00635 \) - **Cross-sectional area (A):** \( A = \frac{\pi d^2}{4} \) ### **Fin Equation (Adiabatic Tip):** For a straight fin with an insulated tip: \[ \frac{T(x) - T_\infty}{T_b - T_\infty} = \frac{\cosh[m(L-x)]}{\cosh(mL)} \] Where: - \( T_b \) = base temperature = 160°C - \( T_\infty \) = air temperature = 24°C --- ## **3. Calculate Fin Parameter \( m \):** \[ m = \sqrt{\frac{h P}{k A}} \] - \( h = 12 \) W/m²·°C - \( P = \pi \times .00635 = .01995 \) m - \( A = \frac{\pi}{4} \times (.00635)^2 = 3.17 \times 10^{-5} \) m² - \( k = 52 \) W/m·°C Plug in: \[ m = \sqrt{\frac{12 \times .01995}{52 \times 3.17 \times 10^{-5}}} \] \[ m = \sqrt{\frac{.2394}{.0016484}} \] \[ m = \sqrt{145.3} \approx 12.05 \text{ m}^{-1} \] --- ## **4. Find the Coolest Temperature of the Handle (at the tip):** At the end (tip) of the fin, \( x = L \): \[ \frac{T(L) - T_\infty}{T_b - T_\infty} = \frac{1}{\cosh(mL)} \] Calculate \( mL \): \[ mL = 12.05 \times .2032 = 2.45 \] Calculate \( \cosh(2.45) \): \[ \cosh(2.45) \approx 5.78 \] So, \[ \frac{T_{tip} - 24}{160 - 24} = \frac{1}{5.78} \] \[ T_{tip} - 24 = \frac{136}{5.78} \] \[ T_{tip} - 24 = 23.5 \] \[ T_{tip} = 47.5^\circ \text{C} \] --- ## **5. How Close to the Door Can You Grasp Without Being Burned?** We want \( T(x) = 65^\circ \)C. \[ \frac{T(x) - 24}{136} = \frac{\cosh[m(L-x)]}{\cosh(mL)} \] Set \( T(x) = 65^\circ C \): \[ \frac{65 - 24}{136} = \frac{\cosh[m(L-x)]}{5.78} \] \[ .301 = \frac{\cosh[m(L-x)]}{5.78} \] \[ \cosh[m(L-x)] = 1.74 \] \[ m(L-x) = \cosh^{-1}(1.74) \approx 1.11 \] \[ L-x = \frac{1.11}{m} = \frac{1.11}{12.05} = .092 \text{ m} = 9.2 \text{ cm} \] So, **you must grasp at least 9.2 cm (3.6 in) away from the door** to avoid burns. --- ## **6. How to Improve the Handle Design?** - **Increase handle diameter:** Reduces temperature drop along handle. - **Use lower thermal conductivity material:** Less heat travels to the end. - **Add insulation to handle:** Reduces heat transfer. - **Increase fin length:** Cooler at the tip, but may compromise usability. --- ## **Summary Table** | **Parameter** | **Value** | |-------------------------|----------------------| | Coolest handle temp | **47.5°C** | | Min safe distance | **9.2 cm (3.6 in)** | | Max safe temp (grab) | 65°C | --- ## **Final Answers** ### **Coolest temperature of the handle:** \[ \boxed{47.5^\circ \text{C}} \] ### **Minimum distance from door to avoid burns:** \[ \boxed{9.2\, \text{cm} \ (3.6\, \text{in})} \] ### **How to improve:** - Increase handle diameter, - Use a material with lower thermal conductivity, - Add insulation, - Increase length if practical. --- **Let me know if you need the calculations explained in more detail or help with another step!**