Heat Transfer Example Problems [work] May 2026

Newton’s law of cooling: [ Q = h , A , (T_s - T_\infty) ] [ 600 = h \cdot 0.5 \cdot (80 - 20) ] [ 600 = h \cdot 0.5 \cdot 60 = h \cdot 30 ] [ h = 20 , \text{W/m}^2\text{·K} ]

Now heat flux: [ q = \frac{1100 - 50}{0.8334} = \frac{1050}{0.8334} \approx 1260 , \text{W/m}^2 ] heat transfer example problems

[ R_{cond} = \frac{\ln(0.06/0.05)}{2\pi \cdot 15} = \frac{\ln(1.2)}{94.2478} = \frac{0.1823}{94.2478} = 0.001934 , \text{m·K/W} ] Newton’s law of cooling: [ Q = h

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} ] \text{m²·K/W} ] For black parallel plates

First, compute the thermal resistances per unit area: [ R_A = \frac{0.2}{1.2} = 0.1667 , \text{m²·K/W} ] [ R_B = \frac{0.1}{0.15} = 0.6667 , \text{m²·K/W} ] [ R_{total} = 0.1667 + 0.6667 = 0.8334 , \text{m²·K/W} ]

For black parallel plates, the net radiation is: [ Q = \sigma A (T_1^4 - T_2^4) ] [ Q = 5.67 \times 10^{-8} \cdot 1 \cdot (500^4 - 300^4) ] Compute: ( 500^4 = 6.25 \times 10^{10} ) ( 300^4 = 0.81 \times 10^{10} ) Difference = ( 5.44 \times 10^{10} )

[ L_c = \frac{D}{6} = \frac{0.02}{6} = 0.00333 , \text{m} ] [ Bi = \frac{h L_c}{k} = \frac{20 \cdot 0.00333}{401} \approx 1.66 \times 10^{-4} \ll 0.1 ] Valid.