## R ri

For normal incidence, both angles are 0 and Eq. (2.47) can be combined with Eq. (2.44) to yield

If one medium is air (n = 1.0), then Eq. (2.48) becomes

Similarly, the transmittance, rr (subscript r indicates that only reflection losses are considered), can be calculated from the average transmittance of the two components as follows:

For a glazing system of N covers of the same material, it can be proven that

The transmittance, Ta (subscript a indicates that only absorption losses are considered), can be calculated from

where K is the extinction coefficient, which can vary from 4 m 1 (for low-quality glass) to 32 (for high-quality glass), and L is the thickness of the glass cover.

The transmittance, reflectance, and absorptance of a single cover (by considering both reflection and absorption losses) are given by the following expressions. These expressions are for the perpendicular components of polarization, although the same relations can be used for the parallel components:

Since, for practical collector covers, t0 is seldom less than 0.9 and r is on the order of 0.1, the transmittance of a single cover becomes

The absorptance of a cover can be approximated by neglecting the last term of Eq. (2.52c):

and the reflectance of a single cover could be found (keeping in mind that p = 1 — a — t) as p = Ta (1 - Tr) = Ta - T (2.55)

For a two-cover system of not necessarily same materials, the following equation can be obtained (subscript 1 refers to the outer cover and 2 to the inner one):

t1t2

+

t1t2

1 - plp2 ,

1 - PlP2 ,

tp2t1

tp2 t1

Example 2.12

A solar energy collector uses a single glass cover with a thickness of 4 mm. In the visible solar range, the refraction index of glass, n, is 1.526 and its extinction coefficient K is 32 m_1. Calculate the reflectivity, transmissivity, and absorptivity of the glass sheet for the angle of incidence of 60° and at normal incidence (0°).

Solution

Angle of incidence = 60°

From Eq. (2.44), the refraction angle 02 is calculated as sin sin 0j sin sin(60)

1.526

34.60

From Eq. (2.51), the transmittance can be obtained as

Ta = e^ cos(e^ = cos(34.6)^ = 0.856 From Eqs. (2.45) and (2.46),

tan2(92 - 91) = tan2(34.6 - 60) tan2(92 + 91) tan2(34.6 + 60)

0.185

0.001

 1 - r, 1 - ri2 1 + r| 1 - (r±ra)2
 1 - ri 1 - ri2 1 + ri 1 - (riO2
 1 - 0.185 1 - 0.1852 1 + 0.185 1 - (0.185 X 0.856)2
 1 - 0.001 1 - 0.0012 1 + 0.001 1 - (0.001 X 0.856)2

P = ¿[r± (1 + T«T±) + ri (1 + TaTii )] = 0.5[0.185(1 + 0.856 X 0.681) + 0.001(1 + 0.856 X 0.998)] = 0.147

 (1 - Ta ) ( 1 - ^ + 1 - ri 1 2 11 - r±Ta,