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外文翻译1

外文翻译1
外文翻译1

外文翻译

学院:嘉兴学院

专业:建筑环境与设备工程

班级:建环121

学号: 201251645123

姓名:陈晨

指导教师:阳季春

5

Radiation Transmission through Glazing: Absorbed Radiation

The transmission, reflection, and absorption of solar radiation by the various parts of a solar collector are important 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 refractive 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 initially will be assumed to be independent of wavelength. This is an excellent assumption for glass, the most common solar collector cover material. Some cover materials have significant optical property variations with wavelength, and spectral dependence of properties is considered in Section 5.7. Incident solar radiation is unpolarized (or only slightly polarized). However, polarization considerations are important as radiation becomes partially polarized as it passes through collector covers.

The last sections of this chapter treat the absorption of solar radiation by collectors, collector-storage walls, and rooms on an hourly and on a monthly average basis.

Reviews of important considerations of transmission of solar radiation have been presented by Dietz (1954, 1963) and by Siegel and Howell (2002).

5.1 REFLECTION OF RADIATION

For smooth surfaces Fresnel has derived expressions for the reflection of unpolarized radiation on passing from medium 1 with a refractive index 1n to medium 2 with refractive index 2n :

2212

21sin ()

sin ()

r θθθθ⊥-=+ (5.1.1)

2212

21tan ()

tan ()r θθθθ-=+P (5.1.2) 2

r i r r I r I ⊥+=

=

P

(5.1.3)

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 unpolarized radiation r ⊥, and Equation 5.1.2 represents the parallel component of unpolarized radiation r P . (Parallel and perpendicular refer to the plane defined by the incident beam and the surface normal.) Equation 5.1.3 then gives the reflection of unpolarized radiation as the average of the two components. The angles θ1 and θ2 are related to the indices of refraction by Snell’s law,

Figure 5.1.1 Angles of incidence and refraction in media with

refractive indices n 1 and n 2.

1122sin sin n n θθ= (5.1.4)

Thus if the angle of incidence and refractive indices are known, Equations 5.1.1 through 5.1.4 are sufficient to calculate the reflectance of the single interface

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

2

1212(0)r i I n n r I n n ??-== ?

+??

(5.1.5)

If one medium is air (i.e., a refractive index of nearly unity), Equation 5.1.5 becomes

2

1(0)1n r n -??= ?

+??

(5.1.6)

Example 5.1.1

Calculate 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.

Solution

At normal incidence, Equation 5.1.6 can be used:

2

0.526(0)0.04342.526r ??== ???

At an incidence angle of 60?, Equation 5.1.4 gives the refraction angle θ2 :

12sin 60sin 34.581.526θ-??

==

???

From Equation 5.1.3, the reflectance is

2222

1sin (25.42)tan (25.42)1

(60)(0.1850.001)0.0932sin (92.58)tan (94.58)2

r ??--=+=+=????

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 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.

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-r ⊥) of the incident

beam reaches the second interface. Of this, 21r ⊥-()

passes through the interface and r ⊥(1-r ⊥)is reflected back to the first, and so on. Summing the transmitted terms, the transmittance for the perpendicular component of polarization is

22

22

(1)1(1)

11n

n r r r r

r r τ∞

⊥⊥

⊥⊥⊥

=⊥⊥

--=-==-+∑ (5.1.7) Exactly the same expansion results when the parallel component of polarization is considered. The components r ⊥and r P are not equal (except at normal incidence), and the transmittance of initially unpolarized radiation is the average transmittance of the two components,

11121(21)1(21)rN

r r N r N r τ⊥⊥?

?--=+ ? ?+-+-??

P P (5.1.8) where the subscript r is a reminder that only reflection losses have been considered. For a system of N covers all of the same materials, a similar analysis yields

11121(21)1(21)rN

r r N r N r τ⊥⊥?

?--=+ ? ?+-+-??

P P (5.1.9)

Figure 5.1.2 Transmission through one nonabsorbing cover.

Example 5.1.2

Calculate the transmittance of two covers of nonabsorbing glass at normal incidence and at 60°. Solution

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, the transmittance is

10.0434

(0)0.8513(0.0434)

r τ-=

=+

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 transmittance is

110.18510.001(60)0.76213(0.185)13(0.001)r τ??

--=+=??++??

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 illustrated 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).

The index 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 losses similar to Figure 5.1.3.

Figure 5.1.3 Transmittance of 1, 2, 3, and 4 nonabsorbing covers having an index of refraction of

1.526.

Table 5.1.1 Average Refractive Index n in Solar

Spectrum of Some Cover Materials

5.2ABSORPTION BY GLAZING

The absorption of radiation in a partially transparent 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:

d I I K d x =- (5.2.1)

where K is the proportionality constant, the extinction coefficient, which is assumed to be a constant in the solar spectrum. Integrating along the actual pathlength in the medium (i.e., from zero to L/cos θ2) yields

Cover Material

Average

n

Glass

1.526 Polymethyl methacrylate 1.49 Polyvinylfluoride

1.45 Polyfluorinated ethylene propylene 1.34 Polytetrafluoroethylene 1.37 Polycarbonate 1.60

2=exp cos a I KL I τθ??=

- ???

transmitted

incident (5.2.2)

where the subscript a is a reminder that only absorption losses have been considered. 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.

5.3OPTICAL PROPERTIES OF COVER SYSTEMS

The transmittance, reflectance, and absorptance of a single cover, allowing for both reflection and absorption losses, can be determined 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 τ⊥ , reflectance ρ⊥ , and absorptance α⊥ of the cover are

()2222(1)111()11a a

a a r r r r r r τττττ⊥⊥

⊥⊥⊥

⊥??

---==??

-+-????

(5.3.1) ()

()

()2

22

111a a a r r r r r τρτττ⊥⊥⊥⊥

⊥⊥⊥-=+=+- (5.3.2)

1(1)1a a r r αττ⊥⊥⊥??

-=-

?-?

? (5.3.3)

Similarresultsarefoundfortheparallelcomponentofpolarization.Forincidentunpolarized radiation, the optical properties are found by the average of the two components.

The equation for the transmittance 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 order of 0.1 for practical collector covers. With this simplification and with Equation 5.1.8, the transmittance of a single cover becomes

a r τττ? (5.3.4)

This is a satisfactory relationship for solar collectors with cover materials and angles of practical interest.

The absorptance of a solar collector cover can be approximated by letting the last term in Equation 5.3.3 be unity so that

1a ατ?- (5.3.5)

Although the neglected term in Equation 5.3.3 is larger than the neglected term in Equation 5.3.1, the absorptance is much smaller than the transmittances so that the overall accuracy of the two approximations is essentially the same.

The reflectance of a single cover is then found from ρ = 1 ? α ? τ, so that

(1)a r a ρττττ?-=- (5.3.6)

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 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.

Example 5.3.1

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.

Solution

At an incidence angle of 60?, the extinction coefficient —optical pathlength product is

2320.0023

0.0894cos cos34.58

KL θ?==

where 34.58 is the refraction angle calculated in Example 5.1.1. The transmittance τa from Equation 5.2.2 is then

()exp 0.08940.915a τ=-=

Using the results of Example 5.1.1 and Equation 5.3.1, the transmittance is found by averaging the transmittances for the parallel and perpendicular components of polarization,

22

220.91510.18510.18510.00110.001210.1851(0.1850.915)10.0011(0.0010.915)0.5(0.625+0.912)0.768

τ??????----=+??

? ?+-?+-???????==

The reflectance is found using Equation 5.3.2 for each component of polarization and averaging:

[]

0.50.185(10.9150.625)0.001(10.9150.912)0.5(0.2910.002)0.147

ρ=+?++?=+=

In a similar manner, the absorptance is found using Equation 5.3.3:

10.91510.18510.001210.1850.91510.0010.9150.085(0.981 1.000)0.085

2

α---??

=

+ ?

-?-???=+= Alternate Solution

The approximate equations can also be used to find these properties. From Equations 5.3.4 and 5.1.8 the transmittance is

0.91510.18510.0010.771210.18510.001τ--??

=

+= ?++??

From Equation 5.3.5, the absorptance is 10.9150.085α=-= and the reflectance is then

10.7710.0850.144ρ=--=

Note that even though the incidence angle was large and poor-quality glass was used in this example so that the approximate equations tend to be less accurate, the approximate method and the exact method are essentially in agreement.

Although Equations 5.3.4 through 5.3.6 were derived for a single cover, they apply equally well to identical multiple covers. The quantity

r τ should be evaluated from Equation 5.1.9 and the quantity

a τfrom Equation 5.2.2 with L equal to the total cover system thickness.

Example 5.3.2

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, and the refractive index is 1.526. Solution

For one sheet at normal incidence,

16.10.00230.0370KL =?=

The transmittance a τis given as

[](0)exp 2(0.0370)0.93a τ=-=

The transmittance accounting for reflection, from Example 5.1.2, is 0.85. The total transmittance is then found from Equation 5.3.4:

(0)(0)(0)0.85(0.93)0.79r a τττ===

From Example 5.1.1, when θ1 = 60?, θ2 = 34.57?, and

2(0.0370)(60)exp 0.91cos34.58a τ??

=-

= ???

and the total transmittance (with r τ= 0.76 from Example 5.1.2) becomes

(60)(60)(60)0.76(0.91)0.69r a τττ===

Figure 5.3.1 gives curves of transmittance as a function of angle of incidence for systems of one to four identical covers of three different kinds of glass. These curves were calculated from Equation 5.3.4 and have been checked by experiments (Hottel and Woertz, 1942).

In a multicover system, the ray-tracing technique used 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 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 following equations can also be used.

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 inner cover:

()

12121212112211τττττττρρρρ⊥⊥????????=+=+ ? ?--??

??????

P P (5.3.7)

()

2121221122τρττρτρρρρρττ⊥⊥???

?????=+=+++ ? ?????????

P P (5.3.8)

Figure 5.3.1 Transmittance (considering absorption and reflection) of one, two, three, and four covers for three types of glass.

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 parentheses.

Example 5.3.3

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. The 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. Solution

The optical properties of the glass and plastic covers alone, as calculated from Equations 5.3.1 through

5.3.3, are

Glass: 0.953τ=P , 0.655τ⊥=

0.002ρ=P , 0.302ρ⊥= 0.044α=P , 0.044α⊥=

Plastic: 0.995τ=P , 0.726τ⊥=

0.005ρ=P , 0.274ρ⊥=

0.000α=P , 0.000α⊥=

Equations 5.3.4 through 5.3.6 could have been used with each component of polarization to simplify the calculation of the preceding properties.

The transmittance of the combination is found from Equation 5.3.7:

10.6550.7260.9530.995210.3020.27410.0020.005=0.50.518+0.948=0.733

τ????

=+ ?

-?-??? ()

The reflectance, with the glass first, is found from Equation 5.3.8:

()10.5180.2740.655

0.9480.0050.953

=0.302+0.002

2

0.726

0.995=0.50.430+0.007=0.219

ρ?????

?++ ?

??

The absorptance is then

=10.2190.7730.048α--=

Equations 5.3.7 and 5.3.8 can be used to calculate the transmittance of 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 appropriate 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 wavelength-dependent properties, integration over the wavelength spectrum is necessary (see Section 5.6).

5 透过玻璃的辐射传递:吸收辐射

集热器的性能对太阳能集热器各个部位传递、反射、吸收的太阳辐射起重要决定作用。透射率、反射率、吸收率与材料的辐射入射角,厚度,折射率,消光系数成函数关系。一般来说,覆盖材料的折射率n 和消光系数K 是辐射波长的函数。在这一章节中,所有的参数属性最初都将假设为独立的波长,对于玻璃这种最常见的太阳能集热器覆盖材料这是一个很好的假设。一些其他的有重要光学性质,波长变化和光谱依赖性的覆盖材料将在章节5.7中有详细介绍。太阳的入射辐射都是非偏振的(或略偏振),然而,对偏振的考虑是十分重要的,因为当光通过太阳能集热器时会发生部分偏振现象。

本章的最后部分对一小时的和月平均的集热器、集热蓄热墙、房间吸收的太阳辐射进行说明。 迪茨(1954、1963)、西格尔和豪厄尔(2002)提出了太阳能辐射传递的重要内容。

5.1反射辐射

菲涅尔推导出非偏振光从折射率为1n 的光滑介质1传递到折射率为2n 的光滑介质2中的反射率公式:

2212

21sin ()

sin ()r θθθθ⊥-=+ (5.1.1) 221221tan ()

tan ()r θθθθ-=

+P (5.1.2) 2

r i r r I r I ⊥+==

P

(5.1.3)

在图5.1.1中1θ 和2θ 分别为入射角和折射角。公式5.1.1表达了非偏振光辐射的垂直方向分量r ⊥ 。公式5.1.2表达了非偏振光辐射的水平方向分量r P (入射光束的垂直和水平分量由参考平面和法线来定义)。公式5.1.3阐述了非偏振光入射的两个组成部分的平均量。

图5.1.1 太阳光通过折射率为1n 和2n 不同介质的折射角和入射角。

斯涅尔定律阐述了折射系数与角1θ 和2θ的关系:

1122sin sin n n θθ= (5.1.4)

因此如果已知折射率和入射角通过公式5.1.1和5.1.4可以明确的计算出一个界面的反射率。 当光线垂直入射时,角1θ 和2θ的值都为零时,公式5.1.3和5.1.4可以整合成

2

1212(0)r i I n n r I n n ??

-== ?+??

(5.1.5)

如果只有一个介质就是空气(也就是说,折射率接近1.0)则公式5.1.5就变成了

2

1(0)1n r n -??= ?+??

(5.1.6)

例题5.1.1

分别计算垂直入射和入射角为60°时,直射辐射在玻璃表面的反射率。已知玻璃的平均折射率为1.526 。

解答:当垂直入射时,用公式5.1.6得:

2

0.526(0)0.04342.526r ??== ???

当入射角为60°时,用公式5.1.4可以求出折射角2θ :

12sin 60sin 34.581.526θ-??

==

???

根据公式5.1.3,得出反射率为:

22221sin (25.42)tan (25.42)1

(60)(0.1850.001)0.0932sin (92.58)tan (94.58)2

r ??--=+=+=????

在太阳能应用中,辐射都是通过平板材料或者是薄膜材料传输的,所以当光通过这两个界面

时都会造成反射损失。当入射光偏离法线时,界面的反射辐射与偏振光的各个组分是不同的。因

此,投射辐射和反射辐射会出现部分偏振现象。所以,对每个偏振分量进行处理是有必要的。

忽略界面的覆盖材料的吸收,如图5.1.2所示,只考虑入射瞬间的偏振光的垂直分量。入射光

束的剩余部分(1-r ⊥)到达第二个界面。同理有2

1r ⊥-()

透射到下一个界面,r ⊥(1-r ⊥)反射回到第一个界面。结合传播条件,得出透射率垂直偏振分量的表达式:

22

220

(1)1(1)

11n

n r r r r

r r τ∞

⊥⊥

⊥⊥⊥

=⊥⊥

--=-==

-+∑ (5.1.7) 同样的也适用与偏振光的平行分量。当垂直分量r ⊥ 和平行分量r P 不相等时(除去法线射入),初始的非偏振光辐射的值为两个分量的平均值,

111211r r r r r τ⊥⊥??--=+ ? ?++??

P P (5.1.8) 小标r 强调,只考虑了反射损失。当面盖由N 层相同性质玻璃组成,类似分析可以得到: 11121(21)1(21)rN r r N r N r τ⊥⊥?

?--=+ ? ?+-+-??

P P (5.1.9)

5.1.2 一层界面的透射率(不考虑

吸收)

例题5.1.2

分别计算垂直入射和入射角为60°时,通过两块玻璃表面的透射率。

解答:当垂直入射时,由例题5.1.1得反射比为0.0434。由于两个偏振分量相等,由公式5.1.9得透射率:

10.0434

(0)0.8513(0.0434)

r τ-=

=+

当入射角为60°时,同样由例题5.1.1得两个分量的反射比为0.185和0.001。由公式5.1.9得透射率:

110.18510.001(60)0.76213(0.185)13(0.001)r τ??

--=+=??++??

太阳光谱的平均折射率为1.526,根据例题5.1.1和5.1.2

用同样的方法可以计算出所有不同入

射角对应的穿过不吸收玻璃的透射率。根据表面覆盖的玻璃块数的不同所有数据都在图5.1.3中画出。这一结果由Hottel 和Woertz (1942)重新校核。

折射指数与太阳能集热器覆盖材料的介质有关,见表5.1.1.。该值对应的太阳光谱可以用来计算类似雨图5.1.3的反射损失。

图5.1.3 折射率为1.526,1,2,3,4层玻璃的透射率(不考虑吸收)

表5.1.1 不同覆盖材料的平均折射率

5.2 玻璃吸收

布格定律描述了半透明介质对辐射的吸收,这是基于假设所吸收的辐射量和与介质中局部辐射量和辐射经过介质的距离x 成正比:

dI IKdx =- (5.2.1)

覆盖材料 平均折射率 玻璃 1.526 聚甲基丙烯酸甲酯

1.49 聚氟乙烯 1.45 聚全氟乙丙烯 1.34 聚四氟乙烯 1.37 聚碳酸酯

1.60

K 称为消光系数,在太阳光谱内假设为常数,透明材料的厚度为 L ,沿着介质中的实际路径积分为(从0到

2

cos L

θ)得:

2=exp cos a I KL I τθ??

=- ???

投射

入射 (5.2.2)

下标a 强调单纯吸收作用引起的透射率。对玻璃来说,K 的值大约从41

m -的 “水白”玻璃(出现白色的边缘)到321

m -氧化铁含量高的(绿边)玻璃。

5.3面盖材料的光学特性

通过追踪光线法,类似于公式5.1.7的推导或者是由西格尔和豪厄尔(2002)所提出的净辐射方法来确定一个面盖考虑反射和吸收两种损失的透射率、反射率和吸收率。其垂直偏振分量τ⊥ 、

ρ⊥ 、α⊥ 分别为:

()2222(1)111()11a a

a a r r r r r r τττττ⊥⊥⊥

⊥⊥⊥

⊥??

---==??

-+-????

(5.3.1) ()

()

()2

22

111a a a r r r r r τρτττ⊥⊥⊥⊥

⊥⊥⊥-=+=+- (5.3.2)

1(1)1a a r r αττ⊥⊥⊥??

-=-

?-??

(5.3.3)

相应的平行偏振分量,形式上与垂直的完全一样。对入射的非偏振辐射,两个分量的平均值就为其光学特性。

面盖的a τ值很少小于0.9,界面反射率r 数量级为0.1,公式5.3.1的末项变成1(它是偏振光水平分量的等效),集热器的透射率方程可以简化。根据公式5.1.8可以得出一个盖面的透射率简化公式:

a r τττ? (5.3.4)

这是一个令人满意的关系,符合太阳能集热器和面盖材料的实际利益。

一个面盖的太阳能集热器的吸收率可以近似由公式5.3.3推出:

1a ατ?- (5.3.5)

尽管公式5.3.3省略项比公式5.3.1省略项来得大,由于吸收率数值上比透射率小许多,所以两个近似式的总精度基本相等。

根据1ρατ=-- 可以得到一个面盖的反射率的近似式:

(1)a r a ρττττ?-=- (5.3.6)

通过对公式5.3.4,5.3.6与公式5.3.1 ,5.3.3的比较可以发现,用精确公式时,涉及各个偏振分量,近似公式的有点在于,只要对r τ 一项作偏振计算。例题5.3.1表现了精确和近似两种不同的方法来求解透射率。

例题 5.3.1

已知玻璃消光系数K=321

m - ,厚度L=32mm ,入射角为60°时,用精确和近似的方法求一层面盖的透射率、反射率、吸收率。

精确解法:

入射角为60°时,消光系数――辐射经过介质的距离的积分,

2320.0023

0.0894cos cos34.58

KL θ?==

34.58°由例题5.1.1的计算得到,a τ 由公式5.2.2计算,

()exp 0.08940.915a τ=-=

由例题5.1.1和公式5.3.1可以得到投射率为垂直偏振分量和平行偏振分量的平均值,

22

220.91510.18510.18510.00110.001210.1851(0.1850.915)10.0011(0.0010.915)0.5(0.625+0.912)0.768

τ??????----=+??

? ?+-?+-???????==

根据公式5.3.2得反射率为垂直偏振分量和平行偏振分量的平均值,

[]

0.50.185(10.9150.625)0.001(10.9150.912)0.5(0.2910.002)0.147

ρ=+?++?=+=

用类似的方法,根据公式5.3.3计算吸收率,

10.91510.18510.001210.1850.91510.0010.9150.085(0.981 1.000)0.085

2

α---??

=

+ ?

-?-???=+= 近似解法:

近似解法也要用到这些数据,从公式5.3.4和5.1.8可得出透射率,

0.91510.18510.0010.771210.18510.001τ--??

=

+= ?++??

根据公式5.3.5,吸收率为,10.9150.085α=-= 同理反射率为,10.7710.0850.144ρ=--=

注意,如果入射角很大或者使用劣质玻璃,使用近似方法往往是不准确的,近似方法和精确方法在本质上是一致的。

尽管通过公式5.3.4和5.3.6推导了一层面盖的计算方法,它们同样适用于相同的多层面盖。变量r τ由公式5.1.9求得,变量a τ由公式5.2.2求得,注意把各层的厚度L 加起来。

例题5.3.2

分别计算入射角为0°和60°时入射到两层玻璃盖面的透射率,每层玻璃厚度为2.3mm 。已知玻璃的消光系数为16.11

m - ,,折射率为1.526 。 解答:

入射角为0°时,射入单层盖面:16.10.00230.0370KL =?= 求出透射率a τ :[](0)exp 2(0.0370)0.93a τ=-=

从例题5.1.2得到只考虑反射损失的透射率是0.85。从5.3.4得到总的透射率,

(0)(0)(0)0.85(0.93)0.79r a τττ===

根据例题5.1.1,当160θ=?时,2=34.58θ? ,2(0.0370)(60)exp 0.91cos34.58a τ??

=-= ???

总的透射率为(根据例题5.1.2得0.76r τ=)

(60)(60)(60)0.76(0.91)0.69r a τττ===

图5.3.1为透射率与入射角的函数关系,系统有三种不同的玻璃面盖。这些曲线的计算公式5.3.4已经由(豪特尔和沃尔茨,1942)进过实验验证。

在多层面盖覆盖系统中,可以根据公式5.1.7来用追踪光线法推导相应的公式。威廉(1953)用广义追踪光线法适用于任意数量和类型的面盖。现代辐射传热计算方法也被应用于这些复杂的情况(爱德华兹,1977;西格尔和豪厄尔,2002)。如果面盖种类一致,推荐使用公式5.3.2,虽然也可以使用下面的公式。对一个有两层面盖的系统,如果两层面盖的材料不一样。下面的公式就是计算透射率和反射率的,下标1指的是上面盖,下标2指的是下面盖:

()

12121212112211τττττττρρρρ⊥⊥????????=+=+ ? ?--??

??????

P P (5.3.7)

()

2121221122τρττρτρρρρρττ⊥⊥???

?????=+=+++ ? ?????????

P P (5.3.8)

图5.3.1三种玻璃1,2,3和4层面盖的透射率(考虑吸收和反射)

面盖系统的反射率决定于上层面盖的特性。下标⊥ 和P 适用于括号内的所有变量。 例题5.3.3

计算两层面盖系统的太阳能集热器当入射角为60°时光学特性。上层面盖的K=16.11

m - ,厚度为2.3mm .下层面盖为折射率等于1.45的聚氯乙烯。塑料薄膜足够薄,塑料的辐射吸收可以忽略。

外文翻译

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外文翻译中文版(完整版)

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