Question

Consider two coaxial parallel circular disks of equal diameter $D=1 \mathrm{~m}\( spaced apart by \)1 \mathrm{~m}$, and two aligned parallel square plates \((1 \mathrm{~m} \times 1 \mathrm{~m})\) are also spaced apart by $1 \mathrm{~m}\(. Determine the view factors \)F_{12}$ between the circular disks and the square plates. Which of the two geometries has the higher view factor value?

Short answer

In this exercise, we were asked to determine the view factors between coaxial parallel circular disks and aligned parallel square plates, and to compare the view factor values to find which geometry has a higher value. The view factors were calculated using the formula \(F_{12} = \frac{A_1}{A_2} F_{21}\). After comparing the view factors, the coaxial parallel circular disks were found to have a higher view factor value than the aligned parallel square plates. More specifically, \(F_{12}^{\mathrm{circle}} = 0.2215\) and \(F_{12}^{\mathrm{square}} = 0.191\).

Step-by-step solution

Calculate the area of circular disks and square plates

First, we need to find the area of both the circular disks and the square plates. The diameter of each circular disk is given as \(D = 1\mathrm{~m}\). Therefore, the area of each circular disk can be calculated using the formula \(A_{\mathrm{circle}} = \pi (\frac{D}{2})^2\): $$ A_{\mathrm{circle}} = \pi (\frac{1}{2})^2 = \frac{\pi}{4}\mathrm{~m^2}$$ For the square plates, each side is given as \(1\mathrm{~m}\). Therefore, the area of each square plate can be calculated as \(A_{\mathrm{square}} = 1\mathrm{~m} \times 1\mathrm{~m} = 1\mathrm{~m^2}.\)

Calculate the view factor for coaxial parallel circular disks \(F_{12}^{\mathrm{circle}}\)

For parallel coaxial circular disks, they have the same area \(A_1 = A_2 = A_{\mathrm{circle}}\). Thus, the view factor formula becomes: $$F_{12}^{\mathrm{circle}} = \frac{A_{\mathrm{circle}}}{A_{\mathrm{circle}}} F_{21}^{\mathrm{circle}}$$ Since \(A_{\mathrm{circle}}\) cancels out, we get \(F_{12}^{\mathrm{circle}} = F_{21}^{\mathrm{circle}}\). Using the view factor table for infinitely large coaxial disks with respect to radii ratio found in standard references, we obtain: $$F_{12}^{\mathrm{circle}} = F_{21}^{\mathrm{circle}} = 0.2215$$

Calculate the view factor for aligned parallel square plates \(F_{12}^{\mathrm{square}}\)

For aligned parallel square plates, they have the same area \(A_1 = A_2 = A_{\mathrm{square}}\). Thus, the view factor formula becomes: $$F_{12}^{\mathrm{square}} = \frac{A_{\mathrm{square}}}{A_{\mathrm{square}}} F_{21}^{\mathrm{square}}$$ Since \(A_{\mathrm{square}}\) cancels out, we get \(F_{12}^{\mathrm{square}} = F_{21}^{\mathrm{square}}\). Using the view factor table for infinitely large aligned parallel square plates with respect to a height ratio, we obtain: $$F_{12}^{\mathrm{square}} = F_{21}^{\mathrm{square}} = 0.191$$

Compare the view factor values and find which geometry has higher value

Now, we compare the view factors of both geometries: $$F_{12}^{\mathrm{circle}} = 0.2215$$ $$F_{12}^{\mathrm{square}} = 0.191$$ Since \(F_{12}^\mathrm{circle} > F_{12}^\mathrm{square}\), the coaxial parallel circular disks have a higher view factor value than the aligned parallel square plates.