Chapter 11 FUNDAMENTALS OF THERMAL RADIATION

Chapter 11 Fundamentals of Thermal Radiation11-42 The variation of reflectivity of a surface withwavelength is given. The average reflectivity,emissivity, and absorptivity of the surface are to bedetermined for two source temperatures. Analysis The average reflectivity of this surface forsolar radiation ( T = 5800 K) is determined to beT ( 3 m)(5800 K) = 17400 mK f 0978746 .0.95( T) f ( T) f ( T)1 02 f ( 1f )1 2 ( 0. 35)( 0. 978746) ( 0. 95)( 10. 978746) 0.3621 11 1Noting that this is an opaque surface, 00.353, mAt T = 5800 K: 1 1 10. 362 0.638Repeating calculations for radiation coming from surfaces at T = 300 K,T ( 3 00001685 .m)(300 K) = 900 mK f10.95 (T) (0.35)(0.0001685) (0.95)(1 0.0001685)At T = 300 K: 1 1 10. 95 0.05and 0.05The temperature of the aluminum plate is close to room temperature, and thus emissivity of the plate willbe equal to its absorptivity at room temperature. That is, room 0.05 0.638swhich makes it suitable as a solar collector. ( s 1and room 0 for an ideal solar collector)11-43 The variation of transmissivity of the glass window of a furnace at a specified temperature withwavelength is given. The fraction and the rate of radiation coming from the furnace and transmittedthrough the window are to be determined.Assumptions The window glass behaves as a black body.Analysis The fraction of radiation at wavelengthssmaller than 3 m isT ( 3 m)(1200 K) = 3600 mK f 0.403607The fraction of radiation coming from the furnace andtransmitted through the window is ( T)1 f 2 (1f ) (0.7)(0.403607) (0)(1 0.403607) 0.2833Then the rate of radiation coming from the furnace and transmitted through the window becomes , mG trAT4 0.283(0.250.25m2)(5.6710 0.782 44W/m.K)(1200K) 2076 W11-19