Question

There is a long heated duct that has a semi-circle cross sectional geometry. The view factor (F21) of the curved surface (2) to the flat surface (1) is_________.

A. 0.36
B. 1
C. 0.64
D. 0
E. 0.5

Answer 1

The view factor from the curved surface to the flat surface in a semi-circular duct is 0.5, which means the correct answer is E. 0.5.

Step-by-step explanation:

The student's question is about calculating the view factor (F21) between the curved surface (surface 2) and the flat surface (surface 1) in a semi-circular duct. The view factor is a dimensionless quantity used in radiative heat transfer to quantify the fraction of radiation leaving one surface that strikes another surface. Based on the geometry of the surfaces and the reciprocity and summation rules of view factors, the view factor from a semi-circular surface (the curved surface) to a flat surface (the base of the semi-circle) is 0.5. Therefore, the correct answer is E. 0.5. To elaborate, the reciprocity rule states that A1*F12 = A2*F21, where A1 and A2 are the areas of surfaces 1 and 2 respectively. The summation rule states that the sum of view factors from a surface to all other surfaces including itself must equal 1. For a semi-circular duct, the view factor from the curved surface to the flat surface is half of the view factor from the curved surface to itself and the flat surface combined, hence it is 0.5.