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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 ratioofthe 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 pinK, Fwaunmn : 0.0719 2.1T. = 0.4 ttm x 2900 K = 1000 umK, 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 05
.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 StefanBoltzmann 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 2.5E5 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 03 0.9 52
0.75 “1
ml 51‘
0.15 :12 02 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 urnK, 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 tunK. [ind
Pup 11—118 ): 0.000; for 9.51;. : 3.0 x 400 : 1200 pmK. [ind F(0_ 1915) : 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 tintK. find
110413“) = 0.002134; for Age. = 3.0 x 3000 = 9000 urnK, 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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 Fall '08
 Rothamer
 Heat Transfer

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