Question

Two black parallel rectangles with dimensions $3 \mathrm{ft} \times 5 \mathrm{ft}\( are spaced apart by a distance of \)1 \mathrm{ft}$. The two parallel rectangles are experiencing radiation heat transfer as black surfaces, where the top rectangle receives a total of 180,000 $\mathrm{Btu} / \mathrm{h}$ radiation heat transfer rate from the bottom rectangle. If the top rectangle has a uniform temperature of \(60^{\circ} \mathrm{F}\), determine the temperature of the bottom rectangle.

Short answer

Answer: The temperature of the bottom rectangle is approximately \(125.88^\circ F\).

Step-by-step solution

Define the Stefan-Boltzmann Law for Radiating Surfaces

The Stefan-Boltzmann Law states that the radiation heat transfer rate between two black surfaces depends on their temperatures (\({T_1}\) and \({T_2}\)) and their emissivities (ε). In our case, both surfaces have an emissivity of 1 since they are black. The equation for the radiation heat transfer rate (Q) between two surfaces is given by: $$ Q = A \sigma (T_1^4 - T_2^4) $$ where A is the surface area of one rectangle, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 × 10^{-8} \mathrm{W} / \mathrm{m}^2 \mathrm{K}^4\)), and \(T_1\) and \(T_2\) are the absolute temperatures of the surfaces in Kelvin.

Convert Provided Temperature to Kelvin

We are given the temperature of the top rectangle in Fahrenheit. Convert the temperature to Kelvin and store it as \(T_1\): $$ T_{1(K)} = \frac{5}{9}(T_{1(°F)} - 32) + 273.15 $$ $$ T_{1(K)} = \frac{5}{9}(60 °F - 32) + 273.15 = 288.71 K $$ Note that \(T_{1}\) (top rectangle temperature) is provided in degrees Fahrenheit, which must be converted to Kelvin to be used in the Stefan-Boltzmann equation.

Convert the Radiation Heat Transfer Rate to \(\mathrm{W}\)

The radiation heat transfer rate (Q) between the rectangles is given in \(\mathrm{Btu/h}\). We need to convert this value to the SI unit of Watts (W) to use in our calculations: $$ Q_{(W)} = Q_{(\mathrm{Btu/h})} \times \frac{1055.06\, \mathrm{J}}{1\, \mathrm{Btu}} \times \frac{1\, \mathrm{h}}{3600\, \mathrm{s}} $$ $$ Q_{(W)} = 180,000\, \mathrm{Btu/h} \times \frac{1055.06\, \mathrm{J}}{1\, \mathrm{Btu}} \times \frac{1\, \mathrm{h}}{3600\, \mathrm{s}} = 52981.7\, \mathrm{W} $$

Calculate the Surface Area of One Rectangle

The dimensions of each rectangle are given in feet, but we need to use the SI unit of meters (m) in our calculations. Convert the dimensions to meters and calculate the surface area (A): $$ 3\, \mathrm{ft} \times 5\, \mathrm{ft} \times \frac{1\, \mathrm{m}}{3.28084\, \mathrm{ft}} \times \frac{1\, \mathrm{m}}{3.28084\, \mathrm{ft}} = 0.929\, \mathrm{m}^{2} $$

Determine the Temperature of the Bottom Rectangle

Plug in the known values (Q, A, \(\sigma\), and \(T_1\)) into the Stefan-Boltzmann equation and solve for \(T_2\): $$ 52981.7\, \mathrm{W} = 0.929\, \mathrm{m}^2 (5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^2 \mathrm{K}^4)(T_2^4 - 288.71^4\, \mathrm{K}^4) $$ Solve for \(T_2^4\) : $$ T_2^4 = \frac{52981.7\, \mathrm{W} + 0.929 \, \mathrm{m}^2 (5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^2 \mathrm{K}^4)(288.71^4 \, \mathrm{K}^4)}{0.929\, \mathrm{m}^2(5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^2 \mathrm{K}^4)} $$ $$ T_2^4 = 123116.55 $$ Now, take the fourth root to find \(T_2\) : $$ T_2 = \sqrt[4]{123116.55} = 326.84\, \mathrm{K} $$

Convert the Bottom Rectangle Temperature Back to Fahrenheit

Finally, convert the bottom rectangle's temperature from Kelvin to Fahrenheit: $$ T_{2(°F)} = \frac{9}{5}(T_{2(K)} - 273.15) + 32 $$ $$ T_{2(°F)} = \frac{9}{5}(326.84\, \mathrm{K} - 273.15) + 32 = 125.88^{\circ} \mathrm{F} $$ The temperature of the bottom rectangle is approximately \(125.88^\circ F\).