An opaque surface, 2 m by 2 m, is maintained at 400 K and is exposed to irradiation with G = 1200 W/m2. The irradiation has the spectral distribution of a blackbody at 4000 K. The surface is diffuse and its spectral absorptivity is ai = 0.3 = 0.5 = 0.75 = 0.9 0< < 0.5 um 0.5 um << 1.0 um 1.0 um < 2.0 um > 2.0 um Determine the absorbed radiation, the emissive power, and the net radiation heat transfer.

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akindelemf akindelemf · Answer 1 · 2020-03-01T18:41:06+01:00

Answer:

Absorbed radiation = 600 W/m²

Emission power = 725.56 W/m²

Net radiation heat transfer = 53.04 w

Explanation:

Data:

Surface Thickness = 2m by 2m

Tc = 400 K

T(black) = 4000 K

G = 1200 W/m2

Step 1: get the wavelenght of the max temperature (wein's law)

H(max) x T = 2898

Step 2: Substitute for T when T = 4,000 K

0.5 < 0.724 < 1

Step 3: Calculate for ρ = 0.5

Step 4: Calculate for absorbed radiation, emmision and net radiation.

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