Let's break down the problem: **Problem Statement:** It is required to quench heat oil to about 500°C for quenching purposes. A ladder is used in the furnace. The section of the ladder is 10 mm x 20 mm. The surrounding temperature is at 27°C. For the temperature of the ladder, at 400 mm from oil, to not reach 50°C, the value of the heat transfer coefficient should be ___ W/m²·°C. Take the value of \( k = 205 \) W/m·°C. ### Given Data: - Length from oil, \( L = 400 \) mm = 0.4 m (distance from the hot end) - Section of ladder: 10 mm x 20 mm = 0.01 m x 0.02 m - \( T_{\infty} \) (surrounding temp): 27°C - \( T_o \) (oil temp): 500°C - \( T(x) \leq 50°C \) at \( x = 0.4 \) m - \( k = 205 \) W/m·°C ### Step 1: Area Calculations - Perimeter, \( P = 2 \times (0.01 + 0.02) = 0.06 \) m - Cross-sectional area, \( A = 0.01 \times 0.02 = 0.0002 \) m² ### Step 2: Fin Equation The temperature along a fin is given by: \[ \frac{T(x)-T_\infty}{T_o-T_\infty} = \exp(-mx) \] Where: \[ m = \sqrt{\frac{hP}{kA}} \] We need \( T(x) \leq 50^\circ C \) at \( x = 0.4 \) m \[ \frac{50-27}{500-27} = \exp(-m \cdot 0.4) \] \[ \frac{23}{473} = \exp(-0.4m) \] \[ \ln\left(\frac{23}{473}\right) = -0.4m \] \[ -2.944 = -0.4m \] \[ m = \frac{2.944}{0.4} = 7.36 \] ### Step 3: Substitute the value of \( m \) \[ m = \sqrt{\frac{hP}{kA}} = 7.36 \] \[ \frac{hP}{kA} = (7.36)^2 = 54.17 \] \[ h = \frac{54.17 \cdot kA}{P} \] \[ h = \frac{54.17 \times 205 \times 0.0002}{0.06} \] \[ h = \frac{2.221}{0.06} \] \[ h = 37.02 \] ### **Answer:** The value of the heat transfer coefficient should be **37 W/m²·°C**. So, the correct option is **37**.