The transmission of solar radiation through glass

Of the energy which is incident upon the glass, some is reflected and lost, some is transmitted through the glass, and some is absorbed by the glass as the energy passes through it. This small amount of absorbed energy raises the temperature of the glass, and the glass eventually transmits this heat, by convection, partly to the room and partly to the exterior. The sum of the transmission coefficient (transmissivity), the absorption coefficient (absorptivity) and the reflection coefficient (reflectivity) is unity.

Table 7.5 gives figures for transmissibility and absorption but, in round figures, for angles of incidence between 60° and 0°, ordinary single window glass transmits about 85 per cent of the energy incident upon it. About 6 per cent is absorbed and the remaining 9 per cent is reflected. As the angle of incidence increases beyond 60°, the transmitted radiation falls off to zero, the reflected amount increasing. The absorption figure remains fairly constant at about 6 per cent for angles of incidence up to 80°.

For double glazing, the picture is more complicated but, in approximate terms, only about 90 per cent of what passes through single glazing is transmitted. Thus, the transmitted percentage for double glazing is about 76 per cent of the incident energy.

4 mm ordinary, clear glass*
Transmissivity 0.87	0.87	0.86	0.84	0.79	0.67	0.42
Absorptivity 0.05	0.05	0.06	0.06	0.06	0.06	0.06
6 mm plate or float, clear glass**
Transmissivity 0.84	0.84	0.83	0.80	0.74	0.62	0.38
Absorptivity 0.08	0.08	0.08	0.08	0.10	0.09	0.09

* Reproduced by kind permission from the ASHRAE Handbook (1993b). ** Based on data published by Pilkington Bros (1969).

Table 7.5 Transmissivity and absorptivity for direct solar radiation through glass. (The absorptivity for indirect radition is taken as 0.06, regardless of the angle of incidence. The transmissivity for indirect radiation is taken as 0.79, regardless of the angle of incidence.)

Angle of incidence O O ^1* O O <N O O 50° 60° 70° 80°

EXAMPLE 7.8

For the window and information given in example 7.7 calculate the instantaneous gain from solar radiation to a room, given that the transmission coefficient for direct radition T is 0.85, that the intensity of direct solar radiation on a surface normal to the rays is 832 W rn“2, that the intensity of sky radiation is 43 W nT2, normal to the window, and that the transmission coefficient %’ for sky radiation is 0.80.

Answer

Total transmitted solar radiation is given by

Q = (transmitted direct radiation x sunlit area of glass)

+ (transmitted sky radiation x total area of glass)

= T/sA«m + * x 43 x Atotal By equation (7.6), for a vertical window

/g = / cos a cos n = 832 cos 43°30′ cos 21°

= 832 x 0.677 Q = 0.85 x 832 x 0.677 x 8.05 + 0.8 x 43 x 8.25 = 4138 W

As is seen later, not all of this energy constitutes an immediate load on the air conditioning system.

To summarise the procedure for assessing the instantaneous solar heat transmission through glass:

(1) For the given date, time of day and latitude, establish the solar declination, d, altitude, a, and azimuth, z, either by equations (7.1), (7.3) and (7.4) or by reference to tables.

(2) Determine the intensity of direct radiation, /, on a surface normal to the sun’s rays for
the particular altitude, either by reference to tables or by means of equation (7.9). Apply the correction factor from Table 7.4, if needed.

(3) For the orientation of the window in question, calculate the wall-solar azimuth angle, n.

(4) Using equation (7.7) directly, or the simplified versions, equations (7.5) or (7.6), calculate the component of the direct radiation, /g, normally incident on the surface. If the surface is vertical, /5 is /v, and if it is horizontal, /5 is /h.

(5) For the particular angle of incidence on the glass, determine the transmission factor X. If i is between 0° and 60°, take x’ as 0.80, in the absence of other information.

(6) Establish the dimensions of any shadows cast by external shading, hence, knowing the dimensions of the windows, calculate the sunlit and total areas.

(7) Knowing the altitude of the sun and the month of the year, refer to tables and find the value of the sky radiation normally incident on the glass. See Table 7.7. Take the transmission factor for sky radiation, x, as 0.85, in the absence of other information.

(8) Calculate the direct transmitted radiation by multiplying x, /5, and the sunlit area, and add to this the product of x’, the sky radiation and the total area. This is the total instantaneous solar radiation transmitted through the window.

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