Download Aptitude Crack
img not found img not found

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.



img not found
img

For help Students Orientation
Mcqs Questions

One stop destination for examination, preparation, recruitment, and more. Specially designed online test to solve all your preparation worries. Go wherever you want to and practice whenever you want, using the online test platform.