外文翻譯--透過玻璃的輻射傳遞吸收輻射(節(jié)選）

1、<p><b>　　外文翻譯</b></p><p>　　學(xué)　　院：　 　　</p><p>　　專　　業(yè)：　 建筑環(huán)境與設(shè)備工程　</p><p>　　班　　級(jí)： 　 　</p><p>　　學(xué)　　號(hào)：　 　　 </

2、p><p>　　姓　　名： 　　　　 　</p><p>　　指導(dǎo)教師：　 　　　　　</p><p>　　5 </p><p>　　

3、Radiation Transmission through </p><p>　　Glazing: Absorbed Radiation</p><p>　　The transmission, reflection, and absorption of solar radiation by the various parts of a solar collector are import

4、ant in determining collector performance. The transmittance, reflectance, and absorptance are functions of the incoming radiation, thickness, refractive index, and extinction coefficient of the material. Generally, the r

5、efractive index n and the extinction coefficient K of the cover material are functions of the wavelength of the radiation. However, in this chapter, all properties </p><p>　　The last sections of this chapter

6、 treat the absorption of solar radiation by collectors, collector-storage walls, and rooms on an hourly and on a monthly average basis.</p><p>　　Reviews of important considerations of transmission of solar r

7、adiation have been presented by Dietz (1954, 1963) and by Siegel and Howell (2002).</p><p>　　5.1 REFLECTION OF RADIATION</p><p>　　For smooth surfaces Fresnel has derived expressions for the ref

8、lection of unpolarized radiation on passing from medium 1 with a refractive index to medium 2 with refractive index :</p><p><b>　　(5.1.1)</b></p><p><b>　　(5.1.2）</b><

9、/p><p><b>　　(5.1.3)</b></p><p>　　where θ1 and θ2 are the angles of incidence and refraction, as shown in Figure 5.1.1. Equation 5.1.1 represents the perpendicular component of unpolariz

10、ed radiation , and Equation 5.1.2 represents the parallel component of unpolarized radiation . (Parallel and perpendicular refer to the plane defined by the incident beam and the surface normal.) Equation 5.1.3 then give

11、s the reflection of unpolarized radiation as the average of the two components. The angles θ1 and θ2 are related to the indices </p><p>　　Figure 5.1.1 Angles of incidence and refraction in media with refra

12、ctive indices n1 and n2.</p><p>　　(5.1.4) </p><p>　　Thus if the angle of incidence and refractive indices are known, Equations 5.1.1 through 5.1.4 are sufficient to ca

13、lculate the reflectance of the single interface</p><p>　　For radiation at normal incidence both θ1 and θ2 are zero, and Equations 5.1.3 and 5.1.4 can be combined to yield</p><p><b>　　（5.1.

14、5）</b></p><p>　　If one medium is air (i.e., a refractive index of nearly unity), Equation 5.1.5 becomes</p><p>　　(5.1.6) </p><p>　　Example 5.1.1</p><p>　　Calcul

15、ate the reflectance of one surface of glass at normal incidence and at 60?. The average index of refraction of glass for the solar spectrum is 1.526.</p><p><b>　　Solution</b></p><p>

16、　　At normal incidence, Equation 5.1.6 can be used:</p><p>　　At an incidence angle of 60?, Equation 5.1.4 gives the refraction angle θ2 :</p><p>　　From Equation 5.1.3, the reflectance is</p>

17、;<p>　　In solar applications, the transmission of radiation is through a slab or film of material so there are two interfaces per cover to cause reflection losses. At off-normal incidence, the radiation reflected

18、at an interface is different for each component of polarization, so the transmitted and reflected radiation becomes partially polarized. Consequently, it is necessary to treat each component of polarization separately.&l

19、t;/p><p>　　Neglecting absorption in the cover material shown in Figure 5.1.2 and considering for the moment only the perpendicular component of polarization of the incoming radiation, （1-） of the incident beam

20、reaches the second interface. Of this, passes through the interface and（1-）is reflected back to the first, and so on. Summing the transmitted terms, the transmittance for the perpendicular component of polarization is&

21、lt;/p><p><b>　　(5.1.7)</b></p><p>　　Exactly the same expansion results when the parallel component of polarization is considered. The components and are not equal (except at normal inci

22、dence), and the transmittance of initially unpolarized radiation is the average transmittance of the two components,</p><p><b>　　(5.1.8)</b></p><p>　　where the subscript r is a remin

23、der that only reflection losses have been considered.</p><p>　　For a system of N covers all of the same materials, a similar analysis yields</p><p><b>　　(5.1.9)</b></p><p&

24、gt;　　Figure 5.1.2 Transmission through one nonabsorbing cover.</p><p>　　Example 5.1.2</p><p>　　Calculate the transmittance of two covers of nonabsorbing glass at normal incidence and at 60°

25、;.</p><p><b>　　Solution</b></p><p>　　At normal incidence, the reflectance of one interface from Example 5.1.1 is 0.0434. From Equation 5.1.9, with both polarization components equal,

26、 the transmittance is</p><p>　　Also from Example 5.1.1 but at a 60? incidence angle, the reflectances of one interface for each component of polarization are 0.185 and 0.001. From Equation 5.1.9, the transmi

27、ttance is</p><p>　　The solar transmittance of nonabsorbing glass, having an average refractive index of 1.526 in the solar spectrum, has been calculated for all incidence angles in the same manner illustrate

28、d in Examples 5.1.1 and 5.1.2. The results for from one to four glass covers are given in Figure 5.1.3. This figure is a recalculation of the results presented by Hottel and Woertz (1942).</p><p>　　The index

29、 of refraction of materials that have been considered for solar collector covers are given in Table 5.1.1. The values correspond to the solar spectrum and can be used to calculate the angular dependence of reflection los

30、ses similar to Figure 5.1.3.</p><p>　　Figure 5.1.3Transmittance of 1, 2, 3, and 4 nonabsorbing covers having an index of refraction of</p><p><b>　　1.526.</b></p><p>　　T

31、able 5.1.1 Average Refractive Index n in Solar</p><p>　　Spectrum of Some Cover Materials</p><p>　　5.2ABSORPTION BY GLAZING</p><p>　　The absorption of radiation in a partially tran

32、sparent medium is described by Bouguer’s law, which is based on the assumption that the absorbed radiation is proportional to the local intensity in the medium and the distance x the radiation has traveled in the medium:

33、</p><p><b>　　(5.2.1)</b></p><p>　　where K is the proportionality constant, the extinction coefficient, which is assumed to be a constant in the solar spectrum. Integrating along the

34、actual pathlength in the medium (i.e., from zero to L/cos θ2) yields</p><p><b>　　(5.2.2)</b></p><p>　　where the subscript a is a reminder that only absorption losses have been consid

35、ered. For glass, the value of K varies from approximately 4 m?1 for ‘‘water white’’ glass (which appears white when viewed on the edge) to approximately 32 m?1 for high iron oxide content (greenish cast of edge) glass.&l

36、t;/p><p>　　5.3OPTICAL PROPERTIES OF COVER SYSTEMS</p><p>　　The transmittance, reflectance, and absorptance of a single cover, allowing for both reflection and absorption losses, can be determined

37、either by ray-tracing techniques similar to that used to derive Equation 5.1.7 or by the net radiation method as described by Siegel and Howell (2002). For the perpendicular component of polarization, the transmittance

38、, reflectance , and absorptance of the cover are</p><p><b>　?。?.3.1）</b></p><p><b>　?。?.3.2）</b></p><p><b>　?。?.3.3）</b></p><p>　　S

39、imilarresultsarefoundfortheparallelcomponentofpolarization.Forincidentunpolarized radiation, the optical properties are found by the average of the two components.</p><p>　　The equation for the transmittance

40、 of a collector cover can be simplified by noting that the last term in Equation 5.3.1 (and its equivalent for the parallel component of polarization) is nearly unity, since τa is seldom less than 0.9 and r is on the ord

41、er of 0.1 for practical collector covers. With this simplification and with Equation 5.1.8, the transmittance of a single cover becomes</p><p><b>　　(5.3.4)</b></p><p>　　This is a sat

42、isfactory relationship for solar collectors with cover materials and angles of practical interest.</p><p>　　The absorptance of a solar collector cover can be approximated by letting the last term in Equation

43、 5.3.3 be unity so that</p><p><b>　　(5.3.5)</b></p><p>　　Although the neglected term in Equation 5.3.3 is larger than the neglected term in Equation 5.3.1, the absorptance is much sm

44、aller than the transmittances so that the overall accuracy of the two approximations is essentially the same.</p><p>　　The reflectance of a single cover is then found from ρ = 1 ? α ? τ, so that</p>&

45、lt;p><b>　　(5.3.6)</b></p><p>　　The advantage of Equations 5.3.4 through 5.3.6 over Equations 5.3.1 through 5.3.3 is that polarization is accounted for in the approximate equations through the

46、single term τr rather than by the more complicated expressions for each individual optical property. Example 5.3.1 shows a solution for transmittance by the exact equations and also by the approximate equations.</p>

47、;<p>　　Example 5.3.1</p><p>　　Calculate the transmittance, reflectance, and absorptance of a single glass cover 2.3mm thick at an angle of 60?. The extinction coefficient of the glass is 32 m?1.</p

48、><p><b>　　Solution</b></p><p>　　At an incidence angle of 60?, the extinction coefficient—optical pathlength product is</p><p>　　where 34.58 is the refraction angle calculat

49、ed in Example 5.1.1. The transmittance τa from Equation 5.2.2 is then</p><p>　　Using the results of Example 5.1.1 and Equation 5.3.1, the transmittance is found by averaging the transmittances for the parall

50、el and perpendicular components of polarization,</p><p>　　The reflectance is found using Equation 5.3.2 for each component of polarization and averaging:</p><p>　　In a similar manner, the absorp

51、tance is found using Equation 5.3.3:</p><p>　　Alternate Solution</p><p>　　The approximate equations can also be used to find these properties. From Equations 5.3.4 and 5.1.8 the transmittance is

52、</p><p>　　From Equation 5.3.5, the absorptance is</p><p>　　and the reflectance is then</p><p>　　Note that even though the incidence angle was large and poor-quality glass was used i

53、n this example so that the approximate equations tend to be less accurate, the approximate method and the exact method are essentially in agreement. </p><p>　　Although Equations 5.3.4 through 5.3.6 were deri

54、ved for a single cover, they apply equally well to identical multiple covers. The quantity should be evaluated from Equation 5.1.9 and the quantity from Equation 5.2.2 with L equal to the total cover system thickness.&l

55、t;/p><p>　　Example 5.3.2</p><p>　　Calculate the solar transmittance at incidence angles of zero and 60? for two glass covers each 2.3mm thick. The extinction coefficient of the glass is 16.1 m?1, a

56、nd the refractive index is 1.526.</p><p><b>　　Solution</b></p><p>　　For one sheet at normal incidence,</p><p>　　The transmittance is given as</p><p>　　The t

57、ransmittance accounting for reflection, from Example 5.1.2, is 0.85. The total transmittance is then found from Equation 5.3.4:</p><p>　　From Example 5.1.1, when θ1 = 60?, θ2 = 34.57?, and</p><p&g

58、t;　　and the total transmittance (with= 0.76 from Example 5.1.2) becomes</p><p>　　Figure 5.3.1 gives curves of transmittance as a function of angle of incidence for systems of one to four identical covers of

59、three different kinds of glass. These curves were calculated from Equation 5.3.4 and have been checked by experiments (Hottel and Woertz, 1942).</p><p>　　In a multicover system, the ray-tracing technique use

60、d to develop Equation 5.1.7 can be used to derive the appropriate equations. Whillier (1953) has generalized the ray-tracing method to any number or type of covers, and modern radiation heat transfer calculation methods

61、have also been applied to these complicated situations (e.g., Edwards, 1977; Siegel and Howell, 2002). If the covers are identical, the approximate method illustrated in Example 5.3.2 is recommended, although the followi

62、ng equa</p><p>　　For a two-cover system with covers not necessarily identical the following equations are for transmittance and reflectance, where subscript 1 refers to the top cover and subscript 2 to the i

63、nner cover:</p><p><b>　　(5.3.7)</b></p><p><b>　　(5.3.8) </b></p><p>　　Figure 5.3.1 Transmittance (considering absorption and reflection) of one, two, three,

64、and four covers for three types of glass.</p><p>　　The reflectance of a cover system depends upon which cover is on top. In these equations the subscripts ⊥ and || apply to all terms in the corresponding par

65、entheses.</p><p>　　Example 5.3.3</p><p>　　Calculate the optical properties of a two-cover solar collector at an angle of 60?. The outer cover is glass with K = 16.1 m?1 and thickness of 2.3mm. T

66、he inner cover is polyvinyl fluoride with refractive index equal to 1.45. The plastic film is thin enough so that absorption within the plastic can be neglected.</p><p><b>　　Solution</b></p>

67、;<p>　　The optical properties of the glass and plastic covers alone, as calculated from Equations 5.3.1 through 5.3.3, are</p><p>　　Glass: , </p><p><b>　　, </b></

68、p><p><b>　　, </b></p><p>　　Plastic: , </p><p><b>　　, </b></p><p><b>　　, </b></p><p>　　Equations 5.3.4 throug

69、h 5.3.6 could have been used with each component of polarization to simplify the calculation of the preceding properties.</p><p>　　The transmittance of the combination is found from Equation 5.3.7:</p>

70、<p>　　The reflectance, with the glass first, is found from Equation 5.3.8:</p><p>　　The absorptance is then</p><p>　　Equations 5.3.7 and 5.3.8 can be used to calculate the transmittance o

71、f any number of covers by repeated application. If subscript 1 refers to the properties of a cover system and subscript 2 to the properties of an additional cover placed under the stack, then these equations yield the ap

72、propriate transmittance and reflectance of the new system. The reflectance ρ1 in Equation 5.3.7 is the reflectance of the original cover system from the bottom side. If any of the covers exhibit strong waveleng</p>

73、<p>　　5 透過玻璃的輻射傳遞：吸收輻射</p><p>　　集熱器的性能對(duì)太陽(yáng)能集熱器各個(gè)部位傳遞、反射、吸收的太陽(yáng)輻射起重要決定作用。透射率、反射率、吸收率與材料的輻射入射角，厚度，折射率，消光系數(shù)成函數(shù)關(guān)系。一般來(lái)說，覆蓋材料的折射率n和消光系數(shù)K是輻射波長(zhǎng)的函數(shù)。在這一章節(jié)中，所有的參數(shù)屬性最初都將假設(shè)為獨(dú)立的波長(zhǎng)，對(duì)于玻璃這種最常見的太陽(yáng)能集熱器覆蓋材料這是一個(gè)很好的假設(shè)。一些其他的有

74、重要光學(xué)性質(zhì)，波長(zhǎng)變化和光譜依賴性的覆蓋材料將在章節(jié)5.7中有詳細(xì)介紹。太陽(yáng)的入射輻射都是非偏振的（或略偏振），然而，對(duì)偏振的考慮是十分重要的，因?yàn)楫?dāng)光通過太陽(yáng)能集熱器時(shí)會(huì)發(fā)生部分偏振現(xiàn)象。</p><p>　　本章的最后部分對(duì)一小時(shí)的和月平均的集熱器、集熱蓄熱墻、房間吸收的太陽(yáng)輻射進(jìn)行說明。</p><p>　　迪茨(1954、1963)、西格爾和豪厄爾(2002)提出了太陽(yáng)能輻射傳遞的

75、重要內(nèi)容。</p><p><b>　　5.1反射輻射</b></p><p>　　菲涅爾推導(dǎo)出非偏振光從折射率為 的光滑介質(zhì)1傳遞到折射率為 的光滑介質(zhì)2中的反射率公式：</p><p><b>　?。?.1.1）</b></p><p><b>　?。?.1.2）</b>&

76、lt;/p><p><b>　?。?.1.3）</b></p><p>　　在圖5.1.1中 和 分別為入射角和折射角。公式5.1.1表達(dá)了非偏振光輻射的垂直方向分量 。公式5.1.2表達(dá)了非偏振光輻射的水平方向分量 （入射光束的垂直和水平分量由參考平面和法線來(lái)定義）。公式5.1.3闡述了非偏振光入射的兩個(gè)組成部分的平均量。</p><p>　　圖

77、5.1.1 太陽(yáng)光通過折射率為和不同介質(zhì)的折射角和入射角。</p><p>　　斯涅爾定律闡述了折射系數(shù)與角 和的關(guān)系：</p><p><b>　?。?.1.4）</b></p><p>　　因此如果已知折射率和入射角通過公式5.1.1和5.1.4可以明確的計(jì)算出一個(gè)界面的反射率。</p><p>　　當(dāng)光線垂直入射時(shí)

78、，角 和的值都為零時(shí)，公式5.1.3和5.1.4可以整合成</p><p><b>　?。?.1.5）</b></p><p>　　如果只有一個(gè)介質(zhì)就是空氣（也就是說，折射率接近1.0）則公式5.1.5就變成了</p><p><b>　?。?.1.6）</b></p><p><b>　

79、　例題5.1.1</b></p><p>　　分別計(jì)算垂直入射和入射角為60°時(shí)，直射輻射在玻璃表面的反射率。已知玻璃的平均折射率為1.526 。</p><p>　　解答：當(dāng)垂直入射時(shí)，用公式5.1.6得：</p><p>　　當(dāng)入射角為60°時(shí)，用公式5.1.4可以求出折射角 ：</p><p>　　根據(jù)公

80、式5.1.3，得出反射率為：</p><p>　　在太陽(yáng)能應(yīng)用中，輻射都是通過平板材料或者是薄膜材料傳輸?shù)模援?dāng)光通過這兩個(gè)界面時(shí)都會(huì)造成反射損失。當(dāng)入射光偏離法線時(shí)，界面的反射輻射與偏振光的各個(gè)組分是不同的。因此，投射輻射和反射輻射會(huì)出現(xiàn)部分偏振現(xiàn)象。所以，對(duì)每個(gè)偏振分量進(jìn)行處理是有必要的。</p><p>　　忽略界面的覆蓋材料的吸收，如圖5.1.2所示，只考慮入射瞬間的偏振光的垂直

81、分量。入射光束的剩余部分（1-）到達(dá)第二個(gè)界面。同理有透射到下一個(gè)界面，（1-）反射回到第一個(gè)界面。結(jié)合傳播條件，得出透射率垂直偏振分量的表達(dá)式：</p><p><b>　?。?.1.7）</b></p><p>　　同樣的也適用與偏振光的平行分量。當(dāng)垂直分量 和平行分量 不相等時(shí)（除去法線射入），初始的非偏振光輻射的值為兩個(gè)分量的平均值，</p>&

82、lt;p><b>　　（5.1.8）</b></p><p>　　小標(biāo)r強(qiáng)調(diào)，只考慮了反射損失。當(dāng)面蓋由N層相同性質(zhì)玻璃組成，類似分析可以得到：</p><p><b>　?。?.1.9）</b></p><p>　　圖5.1.2 一層界面的透射率（不考慮吸收）</p><p><b&g

83、t;　　例題5.1.2</b></p><p>　　分別計(jì)算垂直入射和入射角為60°時(shí)，通過兩塊玻璃表面的透射率。</p><p>　　解答：當(dāng)垂直入射時(shí)，由例題5.1.1得反射比為0.0434。由于兩個(gè)偏振分量相等，由公式5.1.9得透射率：</p><p>　　當(dāng)入射角為60°時(shí)，同樣由例題5.1.1得兩個(gè)分量的反射比為0.185

84、和0.001。由公式5.1.9得透射率：</p><p>　　太陽(yáng)光譜的平均折射率為1.526，根據(jù)例題5.1.1和5.1.2用同樣的方法可以計(jì)算出所有不同入射角對(duì)應(yīng)的穿過不吸收玻璃的透射率。根據(jù)表面覆蓋的玻璃塊數(shù)的不同所有數(shù)據(jù)都在圖5.1.3中畫出。這一結(jié)果由Hottel和Woertz（1942）重新校核。</p><p>　　折射指數(shù)與太陽(yáng)能集熱器覆蓋材料的介質(zhì)有關(guān)，見表5.1.1.。

85、該值對(duì)應(yīng)的太陽(yáng)光譜可以用來(lái)計(jì)算類似雨圖5.1.3的反射損失。</p><p>　　圖5.1.3 折射率為1.526，1,2,3,4層玻璃的透射率（不考慮吸收）</p><p>　　表5.1.1 不同覆蓋材料的平均折射率</p><p><b>　　5.2 玻璃吸收</b></p><p>　　布格定律描述了半透明介質(zhì)對(duì)輻

86、射的吸收，這是基于假設(shè)所吸收的輻射量和與介質(zhì)中局部輻射量和輻射經(jīng)過介質(zhì)的距離x成正比：</p><p><b>　?。?.2.1）</b></p><p>　　K稱為消光系數(shù)，在太陽(yáng)光譜內(nèi)假設(shè)為常數(shù)，透明材料的厚度為L(zhǎng)，沿著介質(zhì)中的實(shí)際路徑積分為（從0到）得：</p><p><b>　　（5.2.2） </b><

87、/p><p>　　下標(biāo)a強(qiáng)調(diào)單純吸收作用引起的透射率。對(duì)玻璃來(lái)說，K的值大約從4的 “水白”玻璃（出現(xiàn)白色的邊緣）到32氧化鐵含量高的（綠邊）玻璃。</p><p>　　5.3面蓋材料的光學(xué)特性</p><p>　　通過追蹤光線法，類似于公式5.1.7的推導(dǎo)或者是由西格爾和豪厄爾(2002)所提出的凈輻射方法來(lái)確定一個(gè)面蓋考慮反射和吸收兩種損失的透射率、反射率和吸收率。

88、其垂直偏振分量 、 、 分別為：</p><p><b>　?。?.3.1）</b></p><p><b>　?。?.3.2）</b></p><p><b>　?。?.3.3）</b></p><p>　　相應(yīng)的平行偏振分量，形式上與垂直的完全一樣。對(duì)入射的非偏振輻射，兩個(gè)

89、分量的平均值就為其光學(xué)特性。</p><p>　　面蓋的值很少小于0.9，界面反射率r數(shù)量級(jí)為0.1，公式5.3.1的末項(xiàng)變成1（它是偏振光水平分量的等效），集熱器的透射率方程可以簡(jiǎn)化。根據(jù)公式5.1.8可以得出一個(gè)蓋面的透射率簡(jiǎn)化公式：</p><p><b>　?。?.3.4）</b></p><p>　　這是一個(gè)令人滿意的關(guān)系，符合太陽(yáng)能

90、集熱器和面蓋材料的實(shí)際利益。</p><p>　　一個(gè)面蓋的太陽(yáng)能集熱器的吸收率可以近似由公式5.3.3推出：</p><p><b>　?。?.3.5）</b></p><p>　　盡管公式5.3.3省略項(xiàng)比公式5.3.1省略項(xiàng)來(lái)得大，由于吸收率數(shù)值上比透射率小許多，所以兩個(gè)近似式的總精度基本相等。</p><p>　

91、　根據(jù) 可以得到一個(gè)面蓋的反射率的近似式：</p><p><b>　?。?.3.6）</b></p><p>　　通過對(duì)公式5.3.4，5.3.6與公式5.3.1 ，5.3.3的比較可以發(fā)現(xiàn)，用精確公式時(shí)，涉及各個(gè)偏振分量，近似公式的有點(diǎn)在于，只要對(duì) 一項(xiàng)作偏振計(jì)算。例題5.3.1表現(xiàn)了精確和近似兩種不同的方法來(lái)求解透射率。</p><p>

92、<b>　　例題 5.3.1</b></p><p>　　已知玻璃消光系數(shù)K=32 ，厚度L=32 ，入射角為60°時(shí)，用精確和近似的方法求一層面蓋的透射率、反射率、吸收率。</p><p><b>　　精確解法:</b></p><p>　　入射角為60°時(shí)，消光系數(shù)――輻射經(jīng)過介質(zhì)的距離的積分，&l

93、t;/p><p>　　34.58°由例題5.1.1的計(jì)算得到， 由公式5.2.2計(jì)算，</p><p>　　由例題5.1.1和公式5.3.1可以得到投射率為垂直偏振分量和平行偏振分量的平均值，</p><p>　　根據(jù)公式5.3.2得反射率為垂直偏振分量和平行偏振分量的平均值，</p><p>　　用類似的方法，根據(jù)公式5.3.3計(jì)算吸

94、收率，</p><p><b>　　近似解法：</b></p><p>　　近似解法也要用到這些數(shù)據(jù)，從公式5.3.4和5.1.8可得出透射率，</p><p>　　根據(jù)公式5.3.5，吸收率為，</p><p><b>　　同理反射率為， </b></p><p>　　注意

95、，如果入射角很大或者使用劣質(zhì)玻璃，使用近似方法往往是不準(zhǔn)確的，近似方法和精確方法在本質(zhì)上是一致的。</p><p>　　盡管通過公式5.3.4和5.3.6推導(dǎo)了一層面蓋的計(jì)算方法，它們同樣適用于相同的多層面蓋。變量由公式5.1.9求得，變量由公式5.2.2求得，注意把各層的厚度L加起來(lái)。</p><p><b>　　例題5.3.2</b></p><

96、;p>　　分別計(jì)算入射角為0°和60°時(shí)入射到兩層玻璃蓋面的透射率，每層玻璃厚度為2.3。已知玻璃的消光系數(shù)為16.1 ，,折射率為1.526 。</p><p><b>　　解答：</b></p><p>　　入射角為0°時(shí)，射入單層蓋面：</p><p><b>　　求出透射率 ： </b

97、></p><p>　　從例題5.1.2得到只考慮反射損失的透射率是0.85。從5.3.4得到總的透射率，</p><p>　　根據(jù)例題5.1.1，當(dāng)時(shí)， ，</p><p>　　總的透射率為（根據(jù)例題5.1.2得）</p><p>　　圖5.3.1為透射率與入射角的函數(shù)關(guān)系，系統(tǒng)有三種不同的玻璃面蓋。這些曲線的計(jì)算公式5.3.4已經(jīng)由

98、(豪特爾和沃爾茨，1942)進(jìn)過實(shí)驗(yàn)驗(yàn)證。</p><p>　　在多層面蓋覆蓋系統(tǒng)中，可以根據(jù)公式5.1.7來(lái)用追蹤光線法推導(dǎo)相應(yīng)的公式。威廉（1953）用廣義追蹤光線法適用于任意數(shù)量和類型的面蓋?，F(xiàn)代輻射傳熱計(jì)算方法也被應(yīng)用于這些復(fù)雜的情況(愛德華茲，1977；西格爾和豪厄爾，2002)。如果面蓋種類一致，推薦使用公式5.3.2，雖然也可以使用下面的公式。對(duì)一個(gè)有兩層面蓋的系統(tǒng)，如果兩層面蓋的材料不一樣。下面的

99、公式就是計(jì)算透射率和反射率的，下標(biāo)1指的是上面蓋，下標(biāo)2指的是下面蓋：</p><p><b>　　（5.3.7）</b></p><p><b>　?。?.3.8）</b></p><p>　　圖5.3.1三種玻璃1，2，3和4層面蓋的透射率（考慮吸收和反射）</p><p>　　面蓋系統(tǒng)的反射率

100、決定于上層面蓋的特性。下標(biāo) 和 適用于括號(hào)內(nèi)的所有變量。</p><p><b>　　例題5.3.3 </b></p><p>　　計(jì)算兩層面蓋系統(tǒng)的太陽(yáng)能集熱器當(dāng)入射角為60°時(shí)光學(xué)特性。上層面蓋的K=16.1 ，厚度為2.3 .下層面蓋為折射率等于1.45的聚氯乙烯。塑料薄膜足夠薄，塑料的輻射吸收可以忽略。</p><p><

101、;b>　　解答：</b></p><p>　　根據(jù)公式5.3.1到5.3.3可以得到玻璃、塑料面蓋的光學(xué)特性，</p><p>　　玻璃： , </p><p><b>　　, </b></p><p><b>　　, </b></p><

102、p>　　塑料： , </p><p><b>　　, </b></p><p><b>　　, </b></p><p>　　根據(jù)公式5.3.4到5.3.6可以用近似方法來(lái)計(jì)算偏振光的各個(gè)量，</p><p>　　由公式5.3.7求出透射率為：</p><

103、;p>　　由公式5.3.8求出反射率為：</p><p><b>　　同理得吸收率為：</b></p><p>　　公式5.3.7和5.3.8可用于計(jì)算任意數(shù)量的面蓋的透射率。如果下標(biāo)1表示面蓋系統(tǒng)的特性，下標(biāo)2表示額外面蓋的特性，那么這些公式可以求出新系統(tǒng)的透射率和反射率。如公式5.3.7中的反射率 表示原面蓋系統(tǒng)的底層面蓋的特性。如果任何面蓋表現(xiàn)出與強(qiáng)烈的波