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## Three sets of parallel plates LM, NR and PQ are given in Figures 1, 2 and 3. The view factor FIJ is defined as the fraction of radiation leaving plate I that is intercepted by plate J. Assume that the values of FLM and FNR are 0.8 and 0.4, respectively. The value of FPQ (round off to one decimal place) is ______.

Asked On2019-07-09 07:52:23 by:Gaganpreet-Gandhi

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Download question setAnswers## Correct Option: B

FLM = F(I) + F(II) + F(III)

Where F is shape factor of surface area I, II and III of M with surface area of ‘L’

0.8 = F(I) + F(II) + F(III)

∵ (I) and (III) are symmetrical in shape so, shape factor of these surface will be same i.e. F(I) = F(III)

So, 0.8 = 2F(I) + F(II)... (1)

In this area (II) is same as fig. (I) area II.

i.e. FNR = F(II) = 0.4... (2)

From equation (1),

Here, FPQ = F(II) + F(III)

= 0.4 + 0.2 = 0.6

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