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Mechanical Engineering QUESTION #2434
Question 1
The net radiation heat exchange rate between a gray surface (emissivity \(\varepsilon\), area \(A\), temperature \(T_s\)) and a large enclosure (temperature \(T_{cav}\)) is:
  • \(q_{rad} = \sigma A(T_s^4 - T_{cav}^4)\) (blackbody only)
  • \(q_{rad} = \varepsilon\sigma A(T_s^4 - T_{cav}^4)\)✔️
  • \(q_{rad} = h_r A(T_s^2 - T_{cav}^2)\)
  • \(q_{rad} = \varepsilon\sigma A(T_s - T_{cav})^4\)
Correct Answer Explanation
Net radiation from a gray surface in a large enclosure: \(q_{rad}=\varepsilon\sigma A(T_s^4-T_{cav}^4)\). This can be linearized as \(q_{rad}=h_r A(T_s-T_{cav})\).