sophiagardens227

06.11.2020 •

Physics

3. Now use the Frequency slider to increase the frequency to a high value. Compare the motion of the particles when the frequency is high to the low-frequency motion. How would you describe the difference?

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Ответ:

tiakimberly3

04.10.2020

Physics

The emissivity of the radiation shield is e_3 = 0.580

Explanation:

From the question we are told that

The temperature of the first parallel plate is T_1 = 650K

The temperature of the second parallel plate is T_2 = 400K

The emissivity of first plate is e_1 = 0.6

The emissivity of first plate is e_2 = 0.9

Generally the total radiation heat that is been transferred without the shield is mathematically represented as

$Q_1 = \frac{\sigma (T_1 ^4 - T_2 ^4)}{\frac{1}{e1 } + \frac{1}{e_2} -1 }$

Where $\sigma$ is the Stefan-Boltzmann constant which has a value $5.67 *10^{-8} \ W \cdot m^{-2} \cdot K^{-1}$

Substituting values

$Q_1 = \frac{ 5.67 *10^{-8} (650 ^4 - 400 ^4)}{\frac{1}{0.6 } + \frac{1}{0.9} -1 }$

$Q_1 = 4876.8 \ W/m^2$

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

$Q_2 = \frac{15}{100} * Q_1$

So $Q_2 = \frac{15}{100} * 4876.8$

Q_2 = 731.52 W/m^2

Now this new radiation heat transfer can be mathematically represented as

$Q_2 = \frac{\sigma (T_1 ^4 - T_2 ^4)}{ [\frac{1}{e_1 } + \frac{1}{e_2 } - 1 ] + n [\frac{1}{e_3} + \frac{1}{e_3} -1 ]}$

Where e_3 the emissivity of the radiation shield and n is the number of radiation shield

Substituting values

$731.52 = \frac{5.67 *10^{-8} (650 ^4 - 400 ^4)}{ [\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ]}$

$731.52 = \frac{1.4175*10^{-5}}{ [\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ]}$

$[\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ] = \frac{1.4175*10^{-5}}{731.52}$

$[\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ] = 1.9377*10^{-8}$

$1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ] = 1.9377*10^{-8} - [\frac{1}{0.6} + \frac{1}{0.9 } - 1 ]$

$\frac{1}{e_3} + \frac{1}{e_3} = 1 .7222222416$

e_3 = 0.580

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