assignchap12-4-6 - PROBLEM 12.004 KNOWN Furnace with...

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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 the-tad : 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 non-direct 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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assignchap12-4-6 - PROBLEM 12.004 KNOWN Furnace with...

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