[ R_{total} = 0.03183 + 0.00193 + 0.2653 = 0.2991 , \text{m·K/W} ]
Using conduction through Layer A: [ q = k_A \frac{T_1 - T_2}{L_A} \quad \Rightarrow \quad 1260 = 1.2 \cdot \frac{1100 - T_2}{0.2} ] [ 1260 = 6 \cdot (1100 - T_2) \quad \Rightarrow \quad 210 = 1100 - T_2 ] [ T_2 = 890^\circ\text{C} ] heat transfer example problems
For forced convection of air, ( h \approx 20 ) is reasonable (typical range: 10–100). If this were natural convection, ( h ) would be closer to 5–10. Problem 3: Radiation – Net Heat Exchange between Two Surfaces Scenario: Two parallel black plates (emissivity ( \varepsilon = 1 )) are at ( T_1 = 500 , \text{K} ) and ( T_2 = 300 , \text{K} ). Each has area ( A = 1 , \text{m}^2 ). Find the net radiative heat transfer from plate 1 to plate 2. (Stefan-Boltzmann constant ( \sigma = 5.67 \times 10^{-8} , \text{W/m}^2\text{K}^4 )) [ R_{total} = 0
[ \frac{T(t) - T_\infty}{T_i - T_\infty} = \exp\left(-\frac{h A_s}{\rho V c_p} t\right) ] For a sphere: ( A_s/V = 6/D ). [ \frac{100 - 25}{200 - 25} = \exp\left(-\frac{20 \cdot 6}{8933 \cdot 0.02 \cdot 385} t\right) ] [ \frac{75}{175} = 0.4286 = \exp(-0.001744 \cdot t) ] [ \ln(0.4286) = -0.8473 = -0.001744 , t ] [ t \approx 486 , \text{seconds} , (\approx 8.1 , \text{minutes}) ] Each has area ( A = 1 , \text{m}^2 )