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Gennadij [26K]
3 years ago
12

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.
Physics
1 answer:
Genrish500 [490]3 years ago
8 0

Answer:

T_{surr}=296.289\ K

In Celsius:

T_{surr}=296.289-273\\T_{surr}=23.289^oC

Explanation:

The formula we are going to use is:

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

Where:

ε is the emissivity

σ is the Stefan constant

T_s is the final temperature of surrounding surfaces

T_{surr} is the required temperature

A_s is the are of surrounding surface

Calculating The area:

A_s=(0.4)(0.4)+4(0.4)(0.2)\\A_s=0.48\ m^2

σ= 5.67*10^{-8}\ W/m^2.K^4

ε =0.95

T_s=55+273

T_s=328 K

\dot Q_{rad=100 W

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

In Celsius:

T_{surr}=296.289-273\\T_{surr}=23.289^oC

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