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assignchap12-22-45 - PROBLEM 12.22 KNOW’N: Lamp with...

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Unformatted text preview: PROBLEM 12.22 KNOW’N: Lamp with prescribed filament area and temperature radiates like a blackbody 102900 K when consuming 100 W. FIND: (a) Efficiency of the lamp for providing visible radiation. and (b) Efficiency as a function of filament temperature for the range 1300 to 3300 K. SCHEMATIC: Glass Envelope q We Fisament, 2 mm x 5 mm. r5 = 2900 K qeiec = 100 W ASSUMPTIONS: (l) Filament behaves as a blackbody. {2) Glass envelope transmits all visible radiation incident upon it. ANALYSIS: (a) We define the efficiency ofthe lamp as the ratio-ofthe radiant power within the visible spectrum (04 — 0.7 tun) to the electrical power required to operate the lamp at the prescribed temperature. 0 = Elvis/Cielec- The radiant power for a blackbody within the visible spectrum is given as 4 ~ 4 q“S = F(0.4um —> 0.7pn1)A50T5 = {Esau—Hm) — 1~(0_,0‘4#m) ]A50'TS using Eq. 12.31 to relate the band emission factors. From Table 12.1. find A: T. = 0.7 pm X 2900 K = 2030 pin-K, Fwaunmn : 0.0719 2.1T. = 0.4 ttm x 2900 K = 1000 um-K, noduwm] = 0.0013 The efficiency is then :7 =[0.0719—0.0013]x2(2x10‘3mx5x10‘3m)5.07x10‘8 w/m2 «49900 K)4100W n=5.62wf100w=5.6% < (b) Using the fHT‘Radiatian Exchange Tool. Blackbody Emission Factor. and Eqs. (1} and (2) above. a model was developed to compute and plot 7] as a function of T5. too 60 -E——--——1 20 i m 4 - 33 .9 0-5 .9 E 02 003 E...".'__'_.. 0.04 1000 1500 2000 2500 3000 3500 Filament temperature, T5 (K; Continued PROBLEM 12.22 (Cont.) Note that the efficiency decreases markedly with reduced filament temperature. At 2900 K. r] z 5.6% while at 2345 K. the efficiency decreases by more than a factor of five to r] : COMMENTS: {1) Based upon this analysis, less than (3% of the energy consumed by the lamp operating at 2900 K is converted to visible light. The transmission of the glass envelope will be less than unity. so the efficiency will be less than the calculated value. (2) Most of the energy is absorbed by the glass envelope and then lost to the surroundings by cenvection and radiation. Also, a significant amount of power is conducted to the lamp base and into the lamp base socket. ‘ (3) The IH'l‘ workspace used to generate the above plot is shown below. It Radiation Exchange Tool - Blackbodi.r Band Emission Factor: I’ The blackbody band emission tactor, Figure 1214 and Table 12.1. is 'r' FL1Ts = F_Iambda__T(lambclal ,Ts) 77 Eq 12.30 I! where units are lambda (micrometers. mum} and T (K) 1' The blackbodyr band emission factor, Figure 12.14 and Table 12.1. is ‘r‘ FL2Ts = F_Iambda TtlambdaZ,Ts) i“! E: 12.30 H Efficiency and rate expressions: eta = qvis i‘ qelec 77 Eq. [1) i. ela_pc = eta ' 100 ill Efficiency, % qelec = 100 I! Electrical power‘ W qvis = (FLETs - FLth) ‘ As ' sigma " TsM fl Eq (2) sigma = some if Stefan-Boltzmann con stant. met‘2.K ffAssigned Variables: T5 = 2900 if Filament temperature. K As = 0.005 ' 0.005 if Filament area. are lambdat = 0.4 If Wavelength, mum; lower limit of visible spectrum |ambda2 = 0.7 if Wavelength mum; upper limit of visible spectrum (“Data Browser Results - Part (a): FL1 Ts FLETs eta eta_pc ovis As Ts lambda 1 lambda2 qeieo sigma 0.001771 0.07135 0.07026 7 .026 7.026 2.5E-5 2900 0.4 0.7 100 5.67E—8 ‘r’ PROBLEM 12.45 KNOWN: Small flat plate maintained at 400 K coated with white paint having spectral absor tivit distribution (Figure 12.23) approxtmated as a stairstep function. Enclosure surlaee inatntaine at 3 00 K with prescribed spectral emissivity distribution. FIND: (a) Total emissivity of the enclosure surface. 8“. and (b) Total emissivity. 8. anti absorptivitv. 0t. ofthe surface. SCHEMATIC: Coated plate Enclosed surface 0.96 0-3 0.9 52 0.75 “1 ml 51‘ 0.15 :12 0-2 gr 0 0.4 3.0 Mum) 0 2.0 1.0.011) ASSUMPTIONS: [1) Coated plate with white paint is diffuse and opaque. so that 011 2 £1. (2) Plate is small compared to the enclosure surface. and (3) Enclosure surface is isothermaL diffuse and opaque. ANALYSIS: (0.) The total emissivity ofthe enclosure surface at '1}; : 3000 K follows from Eq. 12.38 which can he expressed in terms of the bond emission factor. 1:10.”), Eq. 12.30, a” : le(0_,lle ) + £2 [1 —F(0_,-4Tu )] = 0.2x0738+.0.9[1—0.738] 2 0.383 < where. from 'l‘able 12.1. with lilies = 2 11m X 3000 K = 6000 urn-K, Fig.1.] = 0.733. (b) The total emissivity of the coated plate at T = 400 K can be expressed as 8 : also—ms ) + “2lac—an.ran—1001+“:ll- Flo—1001 s : 0.75><O+ 0.15[0.002134—0.000]+ 0.96[1— 0.002134] : 0.958 < where. from Table 12.1. the band emission factors are: for MT; : 0.4 X 400 : 160 tun-K. [ind Pup 11—118 ): 0.000; for 9.51;. : 3.0 x 400 : 1200 pm-K. [ind F(0_ 19-1-5) : 0.002134. The total absorptivity for the irradiation due to the enclosure surface at TCg = 3000 K is a : [lilo—1116511" r12 [WU—12TH ) i Fin—AER?» ):|+a3 [1 7 F10 'j'QTes a : 0.75 x0.002134+ 015103000 — 0.002134] + 0.96 [1 — 08900] = 0240 < where. from Table 12.1. the band emission factors are: for 1111.5 2 0.4 X 3000 : 1200 tint-K. find 110413“) = 0.002134; for Age. = 3.0 x 3000 = 9000 urn-K, find F({)_A1T' )= 0.8900. '- .. 85 COMMENTS: (I) In evaluating the total emissivity and absorptivitv. remember that 0 : s.(s.}_.T.) and 0t : 01(01;_. Gr} where T. is the temperature of the surface and GA is the spectral irradiation, which if the surroundings are large and isothermal, Gk: EMU“). Hence, 0: : [IUIA ,Tm ). For the opaque. diffuse surface, on = e A. [2) Note that the coated plate (white paint) has an absorptivity for the 3000 K—enCIOsure surface irradiation ofot : 0.240. You would expect it to be a low value since the coating appears visually.r “white”. [3)'1‘he emissivity of the coated plate is quite high. 8 = 0958. Would you have expected this of a "white paint"? Most paints are oxide systems (high absorptivity at long wavelengths) with pigmentation (controls the “color” and hence absorptivin in the visible and near infrared regions). ...
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assignchap12-22-45 - PROBLEM 12.22 KNOW’N: Lamp with...

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