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Chapter 13: Problem 34

Consider two black coaxial parallel circular disks of equal diameter \(D\) that are spaced apart by a distance \(L\). The top and bottom disks have uniform temperatures of \(500^{\circ} \mathrm{C}\) and \(520^{\circ} \mathrm{C}\), respectively. Determine the radiation heat transfer coefficient \(h_{\text {rad }}\) between the disks if they are spaced apart by \(L=D\).

Short Answer

Expert verified
Answer: The radiation heat transfer coefficient between the disks is approximately 4.00 W/m²·K.

Step by step solution

01

Convert temperatures to Kelvin

To work with the temperatures in the Stefan-Boltzmann law, we need to convert the temperatures from Celsius to Kelvin: \(T_1 = 500 + 273.15 = 773.15 \, \text{K}\) \(T_2 = 520 + 273.15 = 793.15 \, \text{K}\)
02

Calculate the view factor

Since the disks are parallel, coaxial, and have equal diameters, the view factor \(F_{1 \to 2}\) can be determined using the equation for parallel disks of equal diameter: \(F_{1 \to 2} = \frac{1}{2} \left(1 - \frac{1}{\sqrt{1 + \frac{4L^2}{D^2}}} \right)\) Given that \(L = D\), we have: \(F_{1 \to 2} = \frac{1}{2} \left(1 - \frac{1}{\sqrt{1 + \frac{4D^2}{D^2}}} \right) = \frac{1}{2} \left(1 - \frac{1}{\sqrt{1 + 4}} \right) = \frac{1}{2} \left(1 - \frac{1}{\sqrt{5}} \right)\)
03

Calculate the net radiative heat flux

We can now calculate the net radiative heat flux \(q_{1 \to 2}\) between the disks using the Stefan-Boltzmann law, given as: \(q_{1 \to 2} = F_{1 \to 2} \sigma (T_1^4 - T_2^4)\) where \(\sigma \approx 5.67 \times 10^{-8} \, \text{W} \cdot \text{m}^{-2} \cdot \text{K}^{-4}\) is the Stefan-Boltzmann constant. Substitute the values into the equation: \(q_{1 \to 2} = \frac{1}{2} \left(1 - \frac{1}{\sqrt{5}} \right) \times 5.67 \times 10^{-8} (773.15^4 - 793.15^4)\) Calculate the net radiative heat flux: \(q_{1 \to 2} \approx -80.04 \, \text{W/m}^2\) (negative sign indicates heat transfer is from disk 2 to disk 1)
04

Calculate the radiation heat transfer coefficient

Finally, we can determine the radiation heat transfer coefficient \(h_{\text{rad}}\) by dividing the net radiative heat flux by the temperature difference between the disks: \(h_{\text{rad}} = \frac{-q_{1 \to 2}}{T_2 - T_1} = \frac{80.04}{793.15 - 773.15} \approx 4.00 \, \text{W/m}^2 \cdot \text{K}\) Thus, the radiation heat transfer coefficient between the disks is approximately \(4.00 \, \text{W/m}^2 \cdot \text{K}\).

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Most popular questions from this chapter

The base surface of a cubical furnace with a side length of \(3 \mathrm{~m}\) has an emissivity of \(0.80\) and is maintained at \(500 \mathrm{~K}\). If the top and side surfaces also have an emissivity of \(0.80\) and are maintained at \(900 \mathrm{~K}\), the net rate of radiation heat transfer from the top and side surfaces to the bottom surface is (a) \(194 \mathrm{~kW}\) (b) \(233 \mathrm{~kW}\) (c) \(288 \mathrm{~kW}\) (d) \(312 \mathrm{~kW}\) (e) \(242 \mathrm{~kW}\)

A thin aluminum sheet with an emissivity of \(0.15\) on both sides is placed between two very large parallel plates, which are maintained at uniform temperatures \(T_{1}=900 \mathrm{~K}\) and \(T_{2}=650 \mathrm{~K}\) and have emissivities \(\varepsilon_{1}=0.5\) and \(\varepsilon_{2}=0.8\), respectively. Determine the net rate of radiation heat transfer between the two plates per unit surface area of the plates and compare the result with that without the shield.

Consider an equimolar mixture of \(\mathrm{CO}_{2}\) and \(\mathrm{O}_{2}\) gases at \(800 \mathrm{~K}\) and a total pressure of \(0.5\) atm. For a path length of \(1.2 \mathrm{~m}\), determine the emissivity of the gas.

Consider a \(3-\mathrm{m} \times 3-\mathrm{m} \times 3-\mathrm{m}\) cubical furnace. The base surface is black and has a temperature of \(400 \mathrm{~K}\). The radiosities for the top and side surfaces are calculated to be \(7500 \mathrm{~W} / \mathrm{m}^{2}\) and \(3200 \mathrm{~W} / \mathrm{m}^{2}\), respectively. If the temperature of the side surfaces is \(485 \mathrm{~K}\), the emissivity of the side surfaces is (a) \(0.37\) (b) \(0.55\) (c) \(0.63\) (d) \(0.80\) (e) \(0.89\)

Two very large parallel plates are maintained at uniform temperatures of \(T_{1}=1000 \mathrm{~K}\) and \(T_{2}=800 \mathrm{~K}\) and have emissivities of \(\varepsilon_{1}=\varepsilon_{2}=0.5\), respectively. It is desired to reduce the net rate of radiation heat transfer between the two plates to one-fifth by placing thin aluminum sheets with an emissivity of \(0.1\) on both sides between the plates. Determine the number of sheets that need to be inserted.
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