Answers
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.
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