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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 val ues of FL M and FN R ar e 0.8 and 0.4, respectively. The value of FPQ (round off to one decimal place) is _____.
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- 1.6
- 0.6
- 1.9
- 0.16
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),
| F(III) = | = 0.2 | |
| 2 |
Here, FPQ = F(II) + F(III)
= 0.4 + 0.2 = 0.6
