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Heat Transfer

Q1: The thermal radiative flux from a surface of emissivity = 0.4 is 22.68 kW/m2. The approximate surface temperature (K) is
(Stefan-Boltzman constant = 5.67xl0-8 W/m2.K4)

A 1000

B 727

C 800

D 1200

ANS:A - 1000

To find the surface temperature T, we can use the Stefan-Boltzmann law, which relates the radiative flux (F) to the temperature (T) and the emissivity (ε) of the surface: 4F=εσT4 Given:

  • Radiative flux =22.68F=22.68 kW/m²
  • Emissivity =0.4ε=0.4
  • Stefan-Boltzmann constant =5.67×10−8σ=5.67×10−8 W/m²K⁴
We can rearrange the equation to solve for T: T4=εσF​ 4=22.68×1030.4×5.67×10−8T4=0.4×5.67×10−822.68×103​ 4≈1.0041×1011T4≈1.0041×1011 ≈(1.0041×1011)1/4T≈(1.0041×1011)1/4  KT≈727K So, the approximate surface temperature is 727 K727K. Therefore, the correct answer is 727.



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