Two aligned parallel rectangles with dimensions \(6 \mathrm{~m} \times\) \(8 \mathrm{~m}\) are spaced apart by a distance of \(2 \mathrm{~m}\). If the two parallel rectangles are experiencing radiation heat transfer as black surfaces, determine the percentage of change in radiation heat transfer rate when the rectangles are moved \(8 \mathrm{~m}\) apart.

To calculate the total surface area of both rectangles, we simply multiply the dimensions of the rectangles: \(A_1 = 6 \cdot 8 = 48 \text{~m}^2\) Since there are two rectangles, the combined surface area is: \(A_{1\&2} = 2 \cdot A_1 = 2 \cdot 48 = 96 \text{~m}^2\)

The shape factor is a dimensionless number that represents the proportion of radiant energy leaving one surface that arrives at the second surface. In this case, the shape factor can be determined through the following formula: \(F_{1 \rightarrow 2} = 1 - \frac{1}{1 + \frac{A_{1\&2}}{A_1} \frac{d}{L_{1 \times 2}}}\) where \(d\) is the distance between rectangles, and \(L_{1 \times 2}\) represents the length of the emitting surface onto the receiving surface.

The radiation heat transfer rate between two black surfaces can be defined as follows: \(Q = c \cdot A_1 \cdot F_{1 \rightarrow 2} \cdot \Delta Temp\) In this exercise, the temperature difference and the constant \(c\) remain the same, so the only thing affected by the change in distance between rectangles is the shape factor. Thus, we need to calculate the shape factor for both cases. a) When the rectangles are 2m apart: \(d_1 = 2m\) We have: \(F_{1 \rightarrow 2}(2m) = 1 - \frac{1}{1 + \frac{96}{48} \frac{2}{8}} = 1 - \frac{1}{2} = 0.5\) b) When the rectangles are 8m apart: \(d_2 = 8m\) We have: \(F_{1 \rightarrow 2}(8m) = 1 - \frac{1}{1 + \frac{96}{48} \frac{8}{8}} = 1 - \frac{1}{1 + 2} = \frac{2}{3}\) Now, knowing the shape factors for both cases, we can calculate the radiation heat transfer rates. \(Q_1 = c \cdot A_1 \cdot F_{1 \rightarrow 2}(2m) = k \cdot 48 \cdot 0.5\) \(Q_2 = c \cdot A_1 \cdot F_{1 \rightarrow 2}(8m) = k \cdot 48 \cdot \frac{2}{3}\)

Finally, we can find the percentage change between the two radiation heat transfer rates: \(Percentage\,change = \frac{Q_2 - Q_1}{Q_1} \cdot 100\) \(Percentage\,change = \frac{k \cdot 48 \cdot \frac{2}{3} - k \cdot 48 \cdot 0.5}{k \cdot 48 \cdot 0.5} \cdot 100\) \(Percentage\,change = \frac{\frac{2}{3} - 0.5}{0.5} \cdot 100\) The percentage change in radiation heat transfer rate when the rectangles are moved to 8m apart is: \(Percentage\,change = \frac{1}{3} \cdot 100 = 33.33\%\) Thus, the radiation heat transfer rate increases by 33.33% when the distance between the rectangles is increased from 2m to 8m.