Question

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.

Short answer

Answer: The percentage change in radiation heat transfer rate is 0%.

Step-by-step solution

Determine the radiation heat transfer formula

To find the radiation heat transfer rate between two parallel surfaces, we can use the following formula: $$Q = \frac{A_1 F_{1 \to 2} \sigma(T_1^4 - T_2^4)}{1 - \varepsilon_1 + A_1 \varepsilon_1 \frac{1 - \varepsilon_2}{A_2 \varepsilon_2}}$$ Where: \(Q\) is the radiation heat transfer rate, \(A_1\) and \(A_2\) are the areas of the two surfaces, \(F_{1 \to 2}\) is the view factor from surface 1 to surface 2, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W/m^2K^4}\)), \(T_1\) and \(T_2\) are the temperatures of the two surfaces, \(\varepsilon_1\) and \(\varepsilon_2\) are the emissivities of the two surfaces. Since both rectangles are black surfaces, their emissivities are equal to 1 (\(\varepsilon_1 = \varepsilon_2 = 1\)).

Calculate the areas and the view factor for the rectangles

The area of each rectangle is given by: $$A = 6 \mathrm{~m} \times 8 \mathrm{~m} = 48 \mathrm{~m^2}$$ Since the two rectangles are parallel and aligned, we have \(A_1 = A_2 = 48 \mathrm{~m^2}\) and the view factor \(F_{1 \to 2}\) is equal to 1.

Simplify the radiation heat transfer formula for our problem

Given that both rectangles are black surfaces (\(\varepsilon_1 = \varepsilon_2 = 1\)) and the view factor is equal to 1, we can simplify the formula: $$Q = \frac{A_1 \sigma(T_1^4 - T_2^4)}{1 - 1 + A_1 (1) \frac{1 - 1}{A_2 (1)}} = A_1 \sigma(T_1^4 - T_2^4)$$

Calculate the percentage change in heat transfer

For our problem, we only need to calculate the percentage change in heat transfer when the distance between the rectangles changes from \(2 \mathrm{~m}\) to \(8 \mathrm{~m}\). Since the temperature and the area of the rectangles do not change, we can use the following formula for the percentage change in heat transfer: $$\Delta \% Q = \frac{Q_{8 \mathrm{~m}} - Q_{2 \mathrm{~m}}}{Q_{2 \mathrm{~m}}} \times 100$$ Where \(\Delta \% Q\) is the percentage change in heat transfer, \(Q_{2 \mathrm{~m}}\) is the initial heat transfer rate, and \(Q_{8 \mathrm{~m}}\) is the heat transfer rate when the rectangles are moved \(8 \mathrm{~m}\) apart. However, since our simplified formula only depends on the area and temperature of the rectangles, which do not change with distance, we can conclude that: $$\Delta \% Q = \frac{Q_{8 \mathrm{~m}} - Q_{2 \mathrm{~m}}}{Q_{2 \mathrm{~m}}} \times 100 = \frac{Q_{2 \mathrm{~m}} - Q_{2 \mathrm{~m}}}{Q_{2 \mathrm{~m}}} \times 100 = 0$$ Thus, the percentage change in radiation heat transfer rate is 0% when the rectangles are moved from \(2 \mathrm{~m}\) to \(8 \mathrm{~m}\) apart.