Answer:
The emissivity of the radiation shield is [tex]e_3 = 0.580[/tex]
Explanation:
From the question we are told that
The temperature of the first parallel plate is [tex]T_1 = 650K[/tex]
The temperature of the second parallel plate is [tex]T_2 = 400K[/tex]
The emissivity of first plate is [tex]e_1 = 0.6[/tex]
The emissivity of first plate is [tex]e_2 = 0.9[/tex]
Generally the total radiation heat that is been transferred without the shield is mathematically represented as
[tex]Q_1 = \frac{\sigma (T_1 ^4 - T_2 ^4)}{\frac{1}{e1 } + \frac{1}{e_2} -1 }[/tex]
Where [tex]\sigma[/tex] is the Stefan-Boltzmann constant which has a value [tex]5.67 *10^{-8} \ W \cdot m^{-2} \cdot K^{-1}[/tex]
Substituting values
[tex]Q_1 = \frac{ 5.67 *10^{-8} (650 ^4 - 400 ^4)}{\frac{1}{0.6 } + \frac{1}{0.9} -1 }[/tex]
[tex]Q_1 = 4876.8 \ W/m^2[/tex]
From the question we are told the that using the radiation shield would reduce the radiation heat transfer by 15%
So the new heat transfer is
[tex]Q_2 = \frac{15}{100} * Q_1[/tex]
So [tex]Q_2 = \frac{15}{100} * 4876.8[/tex]
[tex]Q_2 = 731.52 W/m^2[/tex]
Now this new radiation heat transfer can be mathematically represented as
[tex]Q_2 = \frac{\sigma (T_1 ^4 - T_2 ^4)}{ [\frac{1}{e_1 } + \frac{1}{e_2 } - 1 ] + n [\frac{1}{e_3} + \frac{1}{e_3} -1 ]}[/tex]
Where [tex]e_3[/tex] the emissivity of the radiation shield and n is the number of radiation shield
Substituting values
[tex]731.52 = \frac{5.67 *10^{-8} (650 ^4 - 400 ^4)}{ [\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ]}[/tex]
[tex]731.52 = \frac{1.4175*10^{-5}}{ [\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ]}[/tex]
[tex][\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ] = \frac{1.4175*10^{-5}}{731.52}[/tex]
[tex][\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ] = 1.9377*10^{-8}[/tex]
[tex]1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ] = 1.9377*10^{-8} - [\frac{1}{0.6} + \frac{1}{0.9 } - 1 ][/tex]
[tex]\frac{1}{e_3} + \frac{1}{e_3} = 1 .7222222416[/tex]
[tex]e_3 = 0.580[/tex]
