Question

P5. Window glass less than 3cm thick has a spectral transmission Ta of 0.95 in the range of 30 to 2.5 microns and nearly zero elsewhere. Find the total transmission of the glass for black body radiation at 5800K, and (b) black-room radiation at 300K. Discuss the implications of your results.

Answer 1

0.95 Assumption Spectral distribution of irradiation from black body to glass is proportion to black body emission ie Gat Eba 2:5 (um) For Black Body radiation at 5800K Total transmissivity TE S to Gy dx Gg da O As per assumption made : GA « Enb (6800K) 00 T= TaEa (5800K) di - T2 Exib 5800K) d) Eab(5800k). d Eb (5800K) 0.3 2.5 Schulteon) + 'S to Enterokdant St Eusle.com'da 0 ZA Ex,b5800K) + dit TA. Exb(58005)d) 2.5 0-3 Eb 5800K) 2.5 ( 0.95 Exb(5800K) / 0-3 Eb 5800K ) | 25 Eib 5800K) dx blsook) ad} = 0-95 [ For25) - Fowo >>] 2-5) - FO0-3) = 0.95 Eb (5800K) From Heat transfer data book ,- 0-3 um T=5800K =D T = 1740 umk - F10-12 = 0.0335 do 2-5 um T-5800K = 2T= 14500 um.K → FO-2) = 0.9664.

. T = 0.95 0.9664 -0.03357 = 0-88625 =D 88.625% Hence total transmission = t. Eb (5800K) = 0·88625x 5.68x 10-8x 58004 = no 56966109.21 W/m2. For black room at 300K. i. T= 0-3x 300 = 90 um.k 12T= 2.5 X 300 = 750 um K F(0-7X)=0 F l0912) = 0.000012 .. T= 0.95 [0-000012 - 0] = 114x10-5 :: Total transmission = 114x10-5 X 5.68X10-8 x 3004 = 5-245 x 10-3 W/m2. Implications of Result- ( the transmission very much depends on the source from where irradiation is recieved. je temperature of source plays a vital role. © The results above are valid for an assumption Gpa Ebu but it is not true practically that G2 = Ebad. Ga can be replaced by Ebid only when in numerator and denominator Ga is present.