Question

Consider an infinitely long three-sided triangular enclosure with side lengths \(5 \mathrm{~cm}, 3 \mathrm{~cm}\), and \(4 \mathrm{~cm}\). The view factor from the \(5-\mathrm{cm}\) side to the \(4-\mathrm{cm}\) side is (a) \(0.3\) (b) \(0.4\) (c) \(0.5\) (d) \(0.6\) (e) \(0.7\)

Short answer

Now we can calculate the view factor between side A (5 cm side) and side B (4 cm side): \(F_{A -> B} = \frac{4 cm^2}{5 cm^2 + 4 cm^2} = \frac{4}{9}\) The view factor from the 5 cm side to the 4 cm side in the triangular enclosure is \(\frac{4}{9}\).

Step-by-step solution

Identify the surfaces

Let surface A be the 5 cm side, and surface B be the 4 cm side.

Find the area of the triangle

As the triangular enclosure is a right-angled triangle (3-4-5 triangle), we can find its area using the formula: \(Area = \frac{1}{2} \times base \times height = \frac{1}{2} \times 3 \times 4 = 6 cm^2\)

Use the symmetry method to find the view factor

Since the two sides are perpendicular to each other, their normal vectors are also perpendicular. Hence, \(\cos(\theta_{1}) = \cos(\theta_{2}) = 0\). Using the symmetry method, we can find the view factor without finding the angles and the distance between the differential areas as follows: \(F_{A -> B} = \frac{A_{B}}{A_{A}+A_{B}}\) Where \(A_A\) is the area of surface A (5 cm side), which is equal to \(5 cm \times 1 cm = 5 cm^2\), and \(A_B\) is the area of surface B (4 cm side), which is equal to \(4 cm \times 1 cm = 4 cm^2\).

End of solution