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Consider a sealed 20 cm high electronic box whose base dimensions are 40cm x 40cm placed in a vacuum chamber. The emissivity of the outer surface of the box is 0.95. If the electronic components in the box dissipate a total of 100 W of power and the outer surface temperature of the box is not to exceed 55 C, determine the temperature at which the surrounding surfaces must be kept if this box is to be cooled by radiation alone. Assume the heat transfer from the bottom of the box to the stand to be negligible.

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zainsubhani

Answer:

[tex]T_{surr}=296.289\ K[/tex]

In Celsius:

[tex]T_{surr}=296.289-273\\T_{surr}=23.289^oC[/tex]

Explanation:

The formula we are going to use is:

[tex]\dot Q_{rad}=\epsilon\sigma A_s(T_s^4-T_{surr}^4)[/tex]

Where:

ε is the emissivity

σ is the Stefan constant

[tex]T_s[/tex] is the final temperature of surrounding surfaces

[tex]T_{surr}[/tex] is the required temperature

[tex]A_s[/tex] is the are of surrounding surface

Calculating The area:

[tex]A_s=(0.4)(0.4)+4(0.4)(0.2)\\A_s=0.48\ m^2[/tex]

σ= [tex]5.67*10^{-8}\ W/m^2.K^4[/tex]

ε =0.95

[tex]T_s[/tex]=55+273

[tex]T_s[/tex]=328 K

[tex]\dot Q_{rad[/tex]=100 W

[tex]100=0.95(5.67*10^{-8})(0.48)(328^4-T_{surr}^4)\\3867693926=(328^4-T_{surr}^4)\\T_{surr}^4=7706623130\\T_{surr}=296.289\ K[/tex]

In Celsius:

[tex]T_{surr}=296.289-273\\T_{surr}=23.289^oC[/tex]

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