Subtopic: Calculation of power, torque, fuel consumption, heat load and power losses
Question Statement:
Two very large parallel walls (gray bodies) facing each other have emissivities of 0.5 and 0.7. The view factor between these two walls is _____.
Official Answer Key:
0.40 to 0.42
Step-by-Step Solution & Explanation
For two large parallel gray plates, the *geometric* view factor is 1, but the effective emissivity/interchange factor (commonly referred to in this context as the ‘view factor’ accounting for surface gray-body properties) governing net radiant exchange is:
1/Fₑ = 1/ε₁ + 1/ε₂ − 1 = 1/0.5 + 1/0.7 − 1 = 2 + 1.4286 − 1 = 2.4286
Fₑ = 1/2.4286 = 0.4118. This lies within the accepted range 0.40–0.42.
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