Answer to Question #256754 in Mechanical Engineering for Jnl

Question

Two parallel rectangular surfaces 1m x 2m are opposite to each other at a distance of 4 m. The surfaces are black and at 100 °C and 200 °C, respectively. Calculate the heat exchange by radiation between the two surfaces.

Answer 1

$A_1 =A_2 = 1\textsf{ m}× 2\textsf{ m} = 2\textsf{ m²}\\ T_1 = 100°C= 273 + 100= 373K\\ T_2 = 200°C = 273 + 200= 473K\\ \epsilon_1 = 1\\ \epsilon_2 = 1\\ \sigma = 5.670 × 10^{-8} W/m^2K^4$

$\overline{\epsilon}= \dfrac1{\dfrac1{\epsilon_1}+\dfrac{A_1}{A_2}(\dfrac1{\epsilon_2}-1)}$

$=\dfrac1{\dfrac1{1}+\dfrac{1}{1}(\dfrac1{1}-1)}= \dfrac11 = 1$

$Q_{1,2}=\overline{\epsilon}\sigma A(T_2^4-T_1^4)= 1×5.67× 10^{-8}×(473^4-373^4)= 1740.56\textsf{ W}$