Question

The radiative heat transfer rate per unit area (W/m²) between two plane parallel grey surfaces (emissivity = 0.9) maintained at 400 K and 300 K is (Stefan Boltzman constant σ = 5.67 × 10^-8 W/m²K⁴)
1. 992
2. 812
3. 464
4. 567

Answer 1

Correct Answer - Option 2 : 812

ϵ = 0.9, T₁ = 400 K, T₂ = 300 K

σ = 5.67 × 10^-8 W/m²K⁴

Heat transfer per unit area between parallel grey surfaces is given by

\(Q = \frac{{\sigma \left( {T_1^4 - T_2^4} \right)}}{{\frac{1}{\epsilon} + \frac{1}{\epsilon} - 1}}\) 

\(Q = \frac{{5.67 \times {{10}^{ - 8}}\left( {{{400}^4} - {{300}^4}} \right)}}{{\frac{1}{{0.9}} + \frac{1}{{0.9}} - 1}}\) 

Q = 811.84 W/m²