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Chapter 13: Problem 83

Give examples of radiation effects that affect human comfort. 13-84 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.Give examples of radiation effects that affect human comfort.

Short Answer

Expert verified
Answer: The aluminum shield significantly reduces the net rate of radiation heat transfer between the two parallel plates. With the shield, the heat transfer is 7,691.9 W/m², while without the shield, the heat transfer is 16,314.8 W/m².

Step by step solution

01

Determine the Spatial Resistance for Radiation

To determine the net rate of radiation heat transfer between the two plates per unit surface area, we first need to find their spatial resistance for radiation. Spatial resistance is given by: \[R_{s} = \dfrac{1}{\varepsilon} - 1\] For plate 1: \<^latexCode╗ \( R_{s1} = \dfrac{1}{\varepsilon_1} - 1 = \dfrac{1}{0.5} - 1 = 1 \) For plate 2: \( R_{s2} = \dfrac{1}{\varepsilon_2} - 1 = \dfrac{1}{0.8} - 1 = 0.25 \) For the shield: \( R_{s\text{shield}} = \dfrac{1}{0.15} - 1 = 5.67 \)
02

Calculate the Net Radiation Heat Transfer with the Shield

To find the net rate of radiation heat transfer between the two plates with the shield, use the formula: \[Q = \dfrac{\sigma(T_{1}^4 - T_{2}^4)}{R_{s1} + R_{s\text{shield}} + R_{s2}}\] where σ is the Stefan-Boltzmann constant, which is \(5.67 \times 10^{-8} \mathrm{W/m^2K^4}\). Plug in the given values to find net radiation heat transfer: \[Q = \dfrac{5.67 \times 10^{-8}(900^4 - 650^4)}{1 + 5.67 + 0.25} = 7691.9 \mathrm{W/m^2}\]
03

Calculate the Net Radiation Heat Transfer without the Shield

To find the net rate of radiation heat transfer between the two plates without the shield, use the same formula without the shield's spatial resistance term: \[Q' = \dfrac{\sigma(T_{1}^4 - T_{2}^4)}{R_{s1} + R_{s2}}\] Plug in the given values: \[Q' = \dfrac{5.67 \times 10^{-8}(900^4 - 650^4)}{1 + 0.25} = 16314.8 \mathrm{W/m^2}\]
04

Compare the Net Radiation Heat Transfer with and without the Shield

Now, we can compare the net rate of radiation heat transfer between the two plates per unit surface area with and without the shield: - With the shield: \(Q = 7691.9 \mathrm{W/m^2}\) - Without the shield: \(Q' = 16314.8 \mathrm{W/m^2}\) From the results, we can see that the aluminum shield significantly reduces the net rate of radiation heat transfer between the two plates.

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