An electronic box that consumes 180 W of power is cooled by a fan blowing air in
ID: 1851782 • Letter: A
Question
An electronic box that consumes 180 W of power is cooled by a fan blowing air into the box enclosure. The dimensions of the electronics box are 15 cm?50 cm ?50 cm, and all surfaces of the box are exposed to the ambient except the base surface. Temperature measurements indicate that the box is at an average temperature of 32 oC when ambient air and the surrounding walls are at 25?C. If the emissivity of the outer surface of the box is 0.85, determine the fraction of the heat lost from the outer surfaces of the electronic box.
An electronic box that consumes 180 W of power is cooled by a fan blowing air into the box enclosure. The dimensions of the electronics box are 15 cm?50 cm ?50 cm, and all surfaces of the box are exposed to the ambient except the base surface. Temperature measurements indicate that the box is at an average temperature of 32 oC when ambient air and the surrounding walls are at 25?C. If the emissivity of the outer surface of the box is 0.85, determine the fraction of the heat lost from the outer surfaces of the electronic box.Explanation / Answer
Total area exposed to heat transfer = 4 * 50 * 15 + 50 * 50....i.e 4 side walls...plus top...
A = 0.55 m^2
Ts = 32 + 273 = 305 K
Tw = 25 + 273 = 298 K
Rate of heat transfer from outer surface = sigma * epsilon * A * (Ts^4 - Tw^4)
Q = 0.85 * 5.67 * 10^-8 * 0.55 * (305^4 - 298 ^4) = 20.34 W
Fraction of heat lost = 20.34 / 180 =0.113
Answer is 19.1 %
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