Question

A furnace is shaped like a long equilateral-triangular duct where the width of each side is \(2 \mathrm{~m}\). Heat is supplied from the base surface, whose emissivity is \(\varepsilon_{1}=0.8\), at a rate of $800 \mathrm{~W} / \mathrm{m}^{2}\( while the side surfaces, whose emissivities are \)0.4$, are maintained at \(600 \mathrm{~K}\). Neglecting the end effects, determine the temperature of the base surface. Can you treat this geometry as a two-surface enclosure?

Short answer

Based on the given information and calculations, the temperature of the base surface of the triangular duct-shaped furnace is approximately 398.7 K.

Step-by-step solution

Identify given information

We are given the following information in the problem: - Width of each side: \(2 \mathrm{~m}\) - Base surface emissivity: \(\varepsilon_{1}=0.8\) - Heat supplied from the base surface: \(800 \mathrm{~W}/\mathrm{m}^{2}\) - Side surface emissivity: \(\varepsilon_{2}=0.4\) - Side surface temperature: \(600 \mathrm{~K}\)

Determine if we can treat the geometry as a two-surface enclosure

To determine if we can treat the geometry as a two-surface enclosure, we need to make sure that the radiation occurs only between two surfaces (the base surface and the side surface). Since we are neglecting the end effects of the triangular duct, we can conclude that radiation occurs only between the base surface and the side surface, and it can be treated as a two-surface enclosure.

Calculate the radiosity of the base surface

Since we know the heat supplied from the base surface and the emissivity, we can calculate the radiosity of the base surface using the following equation: $$J_1 = \frac{q_{1}}{\varepsilon_{1}} + G \ \ $$ Where \(J_1\) is the radiosity of the base surface, \(q_1\) is the heat supplied, \(\varepsilon_1\) is the base surface emissivity, and \(G\) is the irradiation incident on the base surface. Given \(q_1=800 \mathrm{~W}/\mathrm{m}^{2}\) and \(\varepsilon_1 = 0.8\), the radiosity of the base surface can be calculated as: $$J_1 = \frac{800 \mathrm{~W}/\mathrm{m}^{2}}{0.8} = 1000 \mathrm{~W}/\mathrm{m}^{2}$$

Calculate the radiosity of the side surface

Next, we need to calculate the radiosity of the side surface. Since we know the temperature and emissivity of the side surface, we can calculate the radiosity of the side surface using the following equation: $$J_2 = \varepsilon_{2} \sigma T_2^4 + G \ \ $$ Where \(J_2\) is the radiosity of the side surface, \(\varepsilon_2\) is the side surface emissivity, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{~W}/\mathrm{m^2\cdot K^4}\)), and \(T_2\) is the side surface temperature. Given \(\varepsilon_2 = 0.4\), \(T_2=600 \mathrm{~K}\), and \(G=0\), we can calculate the radiosity of the side surface as: $$ J_2 = 0.4 \times 5.67 \times 10 ^{-8} \mathrm{~W}/\mathrm{m^2\cdot K^4} \times (600 \mathrm{~K})^4 = 244.26\, \mathrm{W}/\mathrm{m}^2 $$

Calculate the net exchange factor

Now, we can calculate the net exchange factor F between the base surface and the side surface. For a two-surface enclosure with an equilateral triangular duct shape and neglecting the end effects, the net exchange factor can be expressed as: $$F_{1-2} = \frac{2}{\sqrt{3}\cdot h}$$ Where \(h\) is the width of each side (in our case, \(2\mathrm{~m}\)). Substitute the given values to calculate the net exchange factor: $$F_{1-2} = \frac{2}{\sqrt{3}\cdot 2} = \frac{1}{\sqrt{3}} \approx 0.577$$

Calculate the heat transfer by radiation between the surfaces

We can now calculate the heat transfer by radiation between the base surface and the side surfaces using the following equation: $$Q_{1-2} = AF_{1-2}(J_1-J_2)$$ Where \(Q_{1-2}\) is the heat transfer by radiation between the surfaces, \(A\) is the area of the base surface, and \(F_{1-2}\) is the net exchange factor. Since the problem asks for the temperature of the base surface, we are interested in the heat transfer per unit area. Thus, we can write the equation as: $$q_{1-2} = F_{1-2}(J_1-J_2)$$ Plugging in the values of \(F_{1-2}\), \(J_1\), and \(J_2\) into the equation: $$q_{1-2} = 0.577 \times (1000\, \mathrm{W}/\mathrm{m}^{2} - 244.26\, \mathrm{W}/\mathrm{m}^{2}) = 436.03\, \mathrm{W}/\mathrm{m}^{2}$$

Calculate the temperature of the base surface

Finally, we can calculate the temperature of the base surface using the following equation: $$q_{1} = \varepsilon_{1} \sigma T_1^4$$ Where \(T_1\) is the temperature of the base surface. We found \(q_{1-2}\) earlier, which is also equal to \(q_{1}\). So, we can rewrite the equation and solve for \(T_1\): $$436.03\, \mathrm{W}/\mathrm{m}^{2} = 0.8 \times 5.67 \times 10^{-8}\, \mathrm{W}/\mathrm{m^2\cdot K^4} \times T_1^4$$ $$T_1^4 = \frac{436.03\, \mathrm{W}/\mathrm{m}^{2}}{0.8 \times 5.67 \times 10^{-8}\, \mathrm{W}/\mathrm{m^2\cdot K^4}} = 1.525 \times 10^8\, \mathrm{K}^4$$ Now, take the fourth root of both sides to find the temperature of the base surface: $$T_1 = \sqrt[4]{1.525 \times 10^8\, \mathrm{K}^4} \approx 398.7\, \mathrm{K}$$ So, the temperature of the base surface is approximately \(398.7\, \mathrm{K}\).