Thermodynamics HW Solutions 884

Thermodynamics HW Solutions 884 - Chapter 11 Fundamentals...

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Chapter 11 Fundamentals of Thermal Radiation 11-46 The variation of emissivity of a surface with wavelength is given. The average emissivity and absorptivity of the surface are to be determined for two temperatures. Analysis ( a ) For T = 5800 K: λμ μ λ 1 1 0 99534 T =⎯ ⎯= (5 . m)(5800 K) = 29,000 mK f The average emissivity of this surface is 0.203 = + = + = ) 99534 . 0 1 )( 9 . 0 ( ) 99534 . 0 )( 2 . 0 ( ) 1 ( ) ( 1 1 2 1 ε f f T ( b ) For T = 300 K: 1 1 0 013754 T (5 . m)(300 K) = 1500 mK f 0.5 0.2 ε λ λ , μ m 5 and 0.89 = + = + = ) 013754 . 0 1 )( 9 . 0 ( ) 013754 . 0 )( 2 . 0 ( ) 1 ( ) ( 1 1 2 1 f f T The absorptivities of this surface for radiation coming from sources at 5800 K and 300 K are, from Kirchhoff’s law, α == 0.203 (at 5800 K) 0.89 (at 300 K) 11-47 The variation of absorptivity of a surface with wavelength is given. The average absorptivity, reflectivity, and emissivity of the surface are to be determined at given temperatures.
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This note was uploaded on 01/24/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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