A t = 5-mm -thick sheet of anodized aluminum is used to reject heat in a space power application. The edge of the sheet is attached to a hot source, and the sheet is maintained at nearly isothermal conditions at T = 300 K . The sheet is not subjected to irradiation. (a) Determine the ne radiation heat transfer from both sides of the 200 mm × 200 mm sheet to deep space. (b) An engineer suggests boring 3-mm-diameter holes through the sheet. The holes are spaced 5 mm apart. The interior surfaces of the holes are anodized after they are bored. Determine the net radiation heat transfer from both sides of the sheet to deep space. (c) As an alternative design, the 3-mm-diameter flat-bottomed holes are not bored completely through the sheet but are bored to depths of 2 mm on each side, leaving a 1-mm-thick web of aluminum separating the bottoms of the holes located on opposite sides of the sheet. Determine the net radiation heat transfer from both sides of the sheet to deep space. (d) Compare the ratio of the net radiation heat transfer to the mass of the sheet for the three designs.
A t = 5-mm -thick sheet of anodized aluminum is used to reject heat in a space power application. The edge of the sheet is attached to a hot source, and the sheet is maintained at nearly isothermal conditions at T = 300 K . The sheet is not subjected to irradiation. (a) Determine the ne radiation heat transfer from both sides of the 200 mm × 200 mm sheet to deep space. (b) An engineer suggests boring 3-mm-diameter holes through the sheet. The holes are spaced 5 mm apart. The interior surfaces of the holes are anodized after they are bored. Determine the net radiation heat transfer from both sides of the sheet to deep space. (c) As an alternative design, the 3-mm-diameter flat-bottomed holes are not bored completely through the sheet but are bored to depths of 2 mm on each side, leaving a 1-mm-thick web of aluminum separating the bottoms of the holes located on opposite sides of the sheet. Determine the net radiation heat transfer from both sides of the sheet to deep space. (d) Compare the ratio of the net radiation heat transfer to the mass of the sheet for the three designs.
Solution Summary: The author analyzes the net radiation heat transfer from both sides of the 200mmtimes sheet to deep space. The emissivity of anodized aluminum is eps
A
t
=
5-mm
-thick sheet of anodized aluminum is used to reject heat in a space power application. The edge of the sheet is attached to a hot source, and the sheet is maintained at nearly isothermal conditions at
T
=
300
K
. The sheet is not subjected to irradiation. (a) Determine the ne radiation heat transfer from both sides of the
200
mm
×
200
mm
sheet to deep space. (b) An engineer suggests boring 3-mm-diameter holes through the sheet. The holes are spaced 5 mm apart. The interior surfaces of the holes are anodized after they are bored. Determine the net radiation heat transfer from both sides of the sheet to deep space. (c) As an alternative design, the 3-mm-diameter flat-bottomed holes are not bored completely through the sheet but are bored to depths of 2 mm on each side, leaving a 1-mm-thick web of aluminum separating the bottoms of the holes located on opposite sides of the sheet. Determine the net radiation heat transfer from both sides of the sheet to deep space. (d) Compare the ratio of the net radiation heat transfer to the mass of the sheet for the three designs.
A horizontal opaque flat plate is well insulated on the edges and the lower surface. The top surface has an area of 5 m2, and it experiences uniform irradiation at a rate of 5000 W. The plate absorbs 4000 W of the irradiation, and the surface is losing heat at a rate of 500 W by convection. If the plate maintains a uniform temperature of 350 K, determine the absorptivity, reflectivity, and emissivity of the plate.
The human skin is “selective” when it comes to theabsorption of the solar radiation that strikes it perpendicularly.The skin absorbs only 50 percent of the incident radiationwith wavelengths between l1 = 0.517 mm and l2 =1.552 mm. The radiation with wavelengths shorter than l1and longer than l2 is fully absorbed. The solar surface maybe modeled as a blackbody with effective surface temperatureof 5800 K. Calculate the fraction of the incident solar radiationthat is absorbed by the human skin.
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