Question

For infinite parallel planes with emissivities 0.4 and 0.2, the interchange view factor for a radiation from surface 1 to surface 2 is about

1. 0.4
2. 0.6
3. 0.2
4. 0.15

Answer 1

Correct Answer - Option 4 : 0.15

Concept:

The interchange view factor is \(\frac{1}{{\rm{R}}}\)

where R is the thermal resistance

\({\rm{R}} = \frac{1}{{{e_1}}} + \frac{1}{{{e_2}}} - 1\)

Where e₁ and e₂ are the emissivities of surfaces 1 and 2.

∴ Interchange view factor = \(\frac{{{e_1}{e_2}}}{{{e_1} + {e_2} - {e_1}{e_2}}}\)

Calculation:

Given:

e₁ = 0.4, e₂ = 0.2

Interchange view factor = \(\frac{{{e_1}{e_2}}}{{{e_1} + {e_2} - {e_1}{e_2}}}\)

Interchange view factor = \(\frac{{0.4 \times 0.2}}{{0.4 + 0.2 - 0.4 \times 0.2}}\) = 0.15

So, interchange view factor  for a radiation from surface 1 to surface 2 is about 0.15