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

This question was previously asked in

ISRO Refrigeration and Air Conditioning 2014 Official

Option 4 : 0.15

ISRO Scientist ME 2020 Paper

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80 Questions
240 Marks
90 Mins

**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 e_{1} and e_{2} 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:__

e_{1} = 0.4, e_{2} = 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**