Let's break down the problem: ### **Given Data** - Ladder section: \( 10\, \text{mm} \times 20\, \text{mm} \) - Length: \( 400\, \text{mm} = .4\, \text{m} \) - \( k = 205\, \text{W/m} \cdot ^\circ\text{C} \) - Furnace temp: \( 500^\circ\text{C} \) - Surrounding temp: \( 27^\circ\text{C} \) - Max temp at .4 m: \( 50^\circ\text{C} \) - Find: **Value of heat transfer coefficient \( h \) (W/m²·°C)** --- ### **Step 1: Fin Problem Identification** This is a **fin** (extended surface) problem. Use the temperature distribution for a fin with one end at base temperature and the other end exposed to the surrounding. #### **Fin Equation:** \[ \theta(x) = \frac{T(x) - T_\infty}{T_b - T_\infty} = \frac{\cosh[m(L-x)] + \frac{h}{m k} \sinh[m(L-x)]}{\cosh(mL) + \frac{h}{m k} \sinh(mL)} \] But for a long fin (or if one end is insulated): \[ \frac{T(L) - T_\infty}{T_b - T_\infty} = e^{-mL} \] Where: - \( m = \sqrt{\frac{hP}{kA}} \) - \( P \) = perimeter = \( 2 \times (10 + 20) = 60\, \text{mm} = .06\, \text{m} \) - \( A \) = cross-sectional area = \( 10 \times 20 = 200\, \text{mm}^2 = 2 \times 10^{-4}\, \text{m}^2 \) --- ### **Step 2: Substitute Values** - \( T_\infty = 27^\circ\text{C} \) - \( T_b = 500^\circ\text{C} \) - \( T(L) = 50^\circ\text{C} \) - \( L = .4\, \text{m} \) \[ \frac{50 - 27}{500 - 27} = e^{-m \times .4} \] \[ \frac{23}{473} = e^{-.4m} \] \[ \ln\left(\frac{23}{473}\right) = -.4m \] \[ \ln(.0486) = -.4m \] \[ -3.028 = -.4m \] \[ m = \frac{3.028}{.4} = 7.57 \] --- ### **Step 3: Solve for \( h \)** Recall \( m = \sqrt{\frac{hP}{kA}} \): \[ m^2 = \frac{hP}{kA} \] \[ h = \frac{m^2 kA}{P} \] Substitute values: - \( m = 7.57 \) - \( k = 205\, \text{W/m} \cdot ^\circ\text{C} \) - \( A = 2 \times 10^{-4}\, \text{m}^2 \) - \( P = .06\, \text{m} \) \[ h = \frac{(7.57)^2 \times 205 \times 2 \times 10^{-4}}{.06} \] Calculate step by step: 1. \( (7.57)^2 = 57.33 \) 2. \( 57.33 \times 205 = 11752.65 \) 3. \( 11752.65 \times 2 \times 104} = 2.35053 \) 4. \( \frac{2.35053}{.06} = 39.18 \) --- ### **Step 4: Final Answer** The closest value is **39 W/m²·°C**. --- ## **Correct Option:** **39**