Question
Three sets of parallel plates LM, NR and PQ are given in Figures 1, 2 and 3. The view factor FU 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 ___________.
Answer :
From the figures we can say that view factor between o parallel 1m plates = 0.4 = FNR
View factor between one 1m plate to another two equally inclined plates = FLM - FNR = 0.8 - 0.4 = 0.4
View factor between one 1m plate to and this equally inclined. Plate = 0.4/2 = 0.2
For figure ‘3’ view factor form one 1m plate to another parallel plate and another equally inclined plate is FPQ = 0.4 + 0.2 = 0.6
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