hw6_sol - PROBLEM 12.18 ICN'O‘WN: Solar flux at outer...

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Unformatted text preview: PROBLEM 12.18 ICN'O‘WN: Solar flux at outer edge of earths atmosphere, 1353 Uni-"ml. FIND: [a] Emissive power of sun, Surface temperature of sun, Wavelength of maximum solar emission, (r1) Earth equilibrimn temperature. SCHE MATIC : ;. DELZfiJOYm ASSUMPTIONS: (1) Sun and earth emit as blackbodies. [2) No attenuation of solar radiation enroute to earth. [3) Earth atmosphere has no eH'ect on earth energy balance. ANALYSIS: (a) Applying conservation of energy to the solar energy crossing two eoncentnc spheres. one haung the radius of the sun and the other having the radial distance from the edge of the earth's atmosphere to the center of the sun ’1 r 3. II. D .p N E3[_Jngll=4:r!R§_e—Tell qs. Hence r .1 4[1_5><1011m—0_55x107m{I x1353wsm3 Y a E; =+=6302 X10 EEK-"111‘. < 1.39X109111l (1}) From Eq. 12.2%. the temperature of the sun is .1154 4 6.302x10? Wm3 - I. 5.6?x10‘3 w.-'m3—K4 .' ." E T5 ='.. F3} = 57"?4 K. <1 (c) From Wieu's law, E4]. 123?. the wavelength of maximum emissron 1s "max— .1. - 5,??4K —U.-.Utt1n. (d) From an energy balance on the earth's surface q . N 3 I . Ee [a D; It =qg [erc 4 It. Hence, from Eq. 12.28. ._1_.-'4 I. 1.1.54 2 - 1353W.-’1n3 - I. 4x5_6?><10‘3 w.-'m3 -K4 J ." as T = — 6 -_4o_.' =2?E§K. <5 COMMENTS: The average earth temperature is higher than 2733 K due to the shielding effect of the earth‘ s atmosphere (transparent to solar radiation but not to longer wavelength earth emission). PROBLEM 12.20 KNOWN: Various surface temperatures. FIND: (a) Yt‘J'nr-‘avelength corresponding to maximum emission for each surface. (b) Fraction of solar emiss10n in UV. VIS and IR portions of the spectrum. ASSUMPIIDNS: (1] Spectral distribution of emission from each surface is approximately that of a blackbody. [2} The sun emits as a hlackbody at 5801] K. ANALYSIS: {a} From Wien's law, SEQ. 12.27". the wavelength of maxrmum emission for blackbody rachation is . C3 28916 Jim-K "max = —= —- T T For the prescribed surfaces Hot Cool Surface Sun Tungsten metal Slain meta]. (5800K) (2500K) (1500K) (3053:) (00K) lmufiiun) 0.51] 1.115 1.93 9.51} 43.3 ‘i (13) From Fig. 12.3, the spectral regions associated with each portion of the spectrum are Spectrum “’arelength limits, [Ina 17V [|I.[|I — [L4 VIE [L4 — [LT IR [LT — IUD For T = 5130th and each of the wavelength limtts, fiom Table 12.1 find: mun) 10"' 0.4 0.? 10" _ FIEpm-K] 50 2530 4000 5.0 5: 103 1-13.93 0 0.125 0.491 1 Hence. the fiaction of the solar emission in each portion of the spectrum is: 1-1-0: 0.125 — 0 = 0.125 F013 = 0.491— 0.125 = 0.3150 F121—D.491=D.509. <1 COEDIENIS: (1] Spectral concentration of surface radiation depends strongly on surface temperature. (2] Much of the UV solar radiation is absorbed in the earth's atmosphere. PROBLEM 12.2 KVD‘WN: A diffuse surface of area A] = lO'lm; emits diffusely with total emissn—‘e power E = 5 X ID4 RES-"nil . TEN—B: (a) Rate this ernission is intercepted by small surface of area A] = 5 x 10—- in1 at a prescribed loc ation and orientation. [b] Irradiation G: on A]. and (c) Compute and plot (33 as a function of the separation distance r; for the range 0.25 S r; 5' 1.0 in for zenith angles 8'; = O. 30 and 60°. SCI-[EEIATICE “1 A2 = 5110-4 m2 “2 <—& _,/' 03 = so” If '/ './ /.f “’2 : m _ /. A1=1o-4 m2 _ 0 E1: 5:104 thz -—— 91- 50 ASSUMPTIDNS: (1] Surface Al eniits diffusely. (2] all may be approximated as a differential surface area and that ISLE/1'22 -c.'-c.' 1. ANALYSIS: The rate at which emission from AL is intercepted byA; follows from Eq. 12.5 written on a total rather than spectral basis. 911—33 =1e,1{9-¢}Aiws~91dW2—1— I21) Since the surface .911 is diffuse. it follows from Eq. 12.13 that 1e.1{9,<.=5}=1e._1 =Eiffl - [2 The solid angle subtended by A; with respect to A; is dm3_1: A: coslfiijlle'lrE2 _ [3] Substituting Eqs. {2) and {3) into Eq. [1] with numerical values gives - 4 J :- . . a 2 u E An. 31 x10 ii- _ 5x10 x 30 {314; =—1-A1cosBI-$ =flxtlfl Jrn12 Kcosofiulx m—fos er-j If :5 MI {O.Sn1}' (1162 =isslswfinlsrxtsm0‘5 1113' ]><1.?32x10_3sr=1.3?8><10_3 w. < Cb] From section 12, 3.3. the irradiation is the rate at which radiation is incident upon the surface per unit surface area, s. 1.3?3x10_3W - s = ‘11—’- = —= area-Ty (5) ‘5: _‘ o A; 5x10“4 111‘ _‘ z. (c) Using the ]HT workspace with the foregoing equations. the G3 was computed as a function of the separation distance for selected zenith angles. The results are plotted below. Continued... PROBLEM 12.2 (Cont) Irradnstlcm G? -:W.'I11"? :l For all zenith angles, G; decreases with increasing separation distance r] . Front Eq. (3]. note that d0);.1 and. hence G], vary inversely as the square of the separation distanee. For any fixed separation distance, G: is a mar-tann when 6; = D" and decreases With increasing 6;, proportional to cos 6;. COEDEENTS: {1) For a diffuse surface. the intensity. L. is independent of direction and related to the ermssive power as L_ 2 E511. Note that It has the units of [sr] in this relation. (3] Note that Eq. 12.5 is an important relation for detennimng the radiant power leaving a surface in a prescribed manner. It has been used here on a total rather than spectral basis. (3] Returning to part (b) and referring to figure 12.10, the irradiation on A2 may be expressed as A :9 G2 = 1L2 e059: fl L 111 L 1 Show that the result is G; = 2.?6 REE-"111'. Explain how this expression follows from Eq. (12.15). PROBLEM 12.10 KVDWN: Spectral distribution of E; for a diffuse surface. I'D-"D: (a) Total eniia sit-'e power E, [b] Total intenaitju' assocmed with directions 8 = D" and B = 30°. and Fraction of emissive power leaving the surface in directions I154 3' B 1'5 IE-"Z. SCI-[El L—‘LTIC: 200 E3. {mez-um} 100 D_ I 5 1o 15 20 ’1'” m] ASSITMPTIGNS: {1] Dififilse Efl11'3510fl. ANALYSIS: {:1} From Eq. 11.11 it follows that ‘00 ‘5 ‘11] ‘15 q ‘20 ‘00 E =JG E;_{.1)d£l =JU [m :11 +J5 (100ml +Jm {40:31:11 +J13(100)d.1 +J20m3d1 E = 100 W.-"m; - 11m (10 — 5} pm + 20011'.-"rn:-,le{15 — 10} pm + 100 mm] pm (10—15) pm E = 2000 Wm: <3 For a diffuse emitter. L is 1r55z5'"'6“_' ofB and Er]. 11.14 gives If.“ '3 _ 5 _ 20001.11, m‘ e If R'sr IE. zoiTRV/mz-sr < (:2) Since the rsurface is diffitse, use Eqrs. 12.10 and 12.14. -3.'r -.'r.-'2 . Etc-[I'M _, W2) 2 JD 114 IE. eo-iti' suit-J {16' do E HIE. Arr 2 . -_‘-.'r H 3 E(H’f4_,nft2) _ LA eosti'sinfidfljfl do _1 Emit; O 2” E II _.rr 1 _ n m4 m=i[i{13 _0_?073](2g_0)]=0_50 { E II I COEDEENI'S: {1) Note hour= a spectral integration rm}! be performed in parts. [2) In performing the integration ofpitrt [c]. recognize the significance of the diffuse emission assumption for which the intensity is uniform in all direct: oils. ...
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hw6_sol - PROBLEM 12.18 ICN'O‘WN: Solar flux at outer...

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