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Unformatted text preview: PROBLEM 12.004 ' KNOWN: Furnace with prescribed aperture and emissive power. FIND: (3) Position of gauge such that irradiation is G = 1000 W/mz, (b) Irradiation when gauge is tilted
0,] z 20”, and (C) Compute and plot the gage irradiation, G, as a function of the separation distance, L, for
the range 100 S L S 300 mm and tilt angles of 9d : 0, 20, and 60”. SCHEMATIC:
l I Detector,
L Ad=1.6x‘iO'5 m2
Furnace aperture, Af __ _________________ __/_,%
9d , E = 3.72x105 wrm2 Irradiation,
0:20 mm G=1000WIm2 ASSUMPTIONS: (l) Furnace aperture emits diffusely, (2) Ad << L2. ANALYSIS: (a) The irradiation on the detector area is deﬁned as the power incident on the surface per
unit area of the surface. That is G = qfed /Ad qf—>d e IeAr C059f (Dd—f (1.2)
where qf ad is the radiant power which leaves Afand is intercepted by Ad. From Eqs. 12.2 and 12.5,
wd_f is the solid angle subtended by surface Ad with respect to Ar, CUd_f = Ad COS 9d /L2 . Noting that since the aperture emits diffusely, Ie : E/Tt (see Eq. 12.14), and hence
G = (E/yz)Af cosaf (Ad cosed /L2 )/Ad (4)
Solving for L2 and substituting for the condition 85 = 00 and 9d : 0°,
L2 : EcosQf cased Af fnG. (5)
7 1/2
L:[3.72x105W/m2x§(20x10"3)2mz/erIOOOW/mz] :193mm. < 1101'] 9d — 200, fad 136 reduced b a factor Of COS 9d Sil’lCC (Dd; is reduced 1) El factor COS 85,.
Hence, G = 1000 Wimzxcos e, =1000W/m2 xcos20° : 940 W/mz. < (c) Using the IHT workspace with Eq. (4), G is computed and plotted as a function of L for selected 0d.
Note that G decreases inversely as L2. As expected, G decreases with increasing 0d and in the limit,
approaches zero as 6,, approaches 90°. " 3000 2000 irradiation, G (Wm/‘2) i  i i i
100 200 300
Separation distance. L (mm) —9— thetad = 0 deg thetad : 20 deg
—E— thetad = 60 deg PROBLEM 12.6
KNOWN: Flux and intensity of direct and difﬁase components, respectively, of solar irradiation.
FIN D: Total irradiation.
SC HEMATIC: ﬁeao"
93,, = 1000 W/ma\\«
«k Idif = 70 W/m‘2 sr ANALYSIS: Since the irradiation is based on the actual surface area, the contribution due to the
direct solar radiation is Gdir : Cl’dir “3059 From Eq. 12.19 the contribution due to the diffuse radiation is Gdif = 7171dif 7*
Hence
G = Gdir + Gdif = q’air cost? +7L'Idif
01'
G =1000W/m2 x0.866+7rsrx70W/m2 sr
G=(866+220)W/m2
01' G:1086W/m2. < COMMENTS: Although a diffuse approximation is often made for the nondirect component of solar
radiation, the actual directional distribution deviates ﬁom this condition, providing larger intensities at
angles close to the direct beam. ...
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 Fall '08
 Rothamer
 Solar System, Power, Light, Heat Transfer, Ultraviolet

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