Two long parallel surfaces each of emissivity 0.7 are maintained at different temperatures and accordingly have radiation heat exchange between them. It is desired to reduce 75% of the radiant heat transfer by inserting thin parallel shields of emissivity 1 on both sides. The number of shields should be
A. One
B. Two
C. Three
D. Four
Answer: Option C
Solution (By Examveda Team)
$$\frac{{{\text{Q with shield}}}}{{{\text{Q without shield}}}} = \frac{1}{{{\text{n}} + 1}}.$$N = number of parallel shields
0.25n + 0.25 = 1
0.25n = 0.75
n = 3.
This Question Belongs to Mechanical Engineering >> Heat And Mass Transfer
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