Question

A 3-mm-thick glass window transmits 90 percent of the radiation between ? = 0.3 and 30 ?m and is essentially opaque for radiation at other wavelengths Given: Stefan-Boltzmann constant () 5.67 x 10-8 W/m2 K4 value 10.00 polnts Determine the rate of radiation transmitted through a 2-mx 2-m glass window from a blackbody source at 5880 K The rate of radiation transmitted through a 2-m x 2-m glass window from a blackbody source, Etransmit KW. (Correct to three decimal places.)

Answer 1

Given T = 5880 k Area = 2 times 2 = 4 m^2 lambda _1 = 0.3 mu m lambda _2 = 3 mu m. for Black source at 5880 K, so Total Black body Radiation. E_b = sigma T^4 A = (5.67 times 10^- 8) times (5880)^4 times 4 = 271.114 times 10^6 W The Fraction of Radiation in Range 0.3 to 3 mu m lambda _2 T = (0.3) times (5880) = 1764 From Black body Radiation Function fp_lambda Given by lambda _1 T = 0.3 times 5880 4m middot K = 1764 mu mk f_ lambda_1 = using interpolation method. F_ lambda _1 = 0.019718 + (1764 - 1600/1800 - 1600) (0.039341 - 0.019716) f_ lambda _1 = 0.035808 and lambda _2 T = 3 times 5880 = 17640 mu mk. f lambda _2 = 0.973814 + (17640 - 16000/18000 - 16000) (0.980860 - 0.973814) f lambda _2 = 0.979591 delta f = 0.979591 - 0.035808 = 0.9437837

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