# Minor factors influencing solar gains

There are five relatively minor factors that should be taken into account when calculating the instantaneous heat transmission resulting from solar radiation:

(1) |
Atmospheric haze |

(2) |
The type of window frame |

(3) |
The height of the place above sea level |

(4) |
Variation of the dew point |

(5) |
The hemisphere |

(1) |
Atmospheric haze |

This is most noticeable in industrial areas and, in this context, is regarded as resulting from contaminants emitted by traffic and industry, carried aloft by thermal up-currents. It is usually more pronounced in the afternoon, owing to the build up of the ground surface temperatures which produce the up-currents as the day progresses. Haze can reduce the value of /s, the direct radiation normally incident on a surface, so much that a diminution of 15 per cent in the total radiation received (scattered plus direct) may occur. A conservative factor of 0.95 may be applied for cities such as London.

(2) The type of window frame

Tables of the actual heat gain occurring through windows are often published (see Table 7.9). Generally, tables of this sort are for wooden-framed windows. If, as in many modern buildings, the framework is metal, then because this is a much better thermal conductor than wood, the heat gains are increased by about 17 per cent.

A more convenient way of taking account of this effect is to apply the tabulated figure for the heat gain to the glass area, in wooden-framed windows, and to the area of the opening in the wall, in metal-framed windows.

(3) The height of the place above sea level

There is some difference of opinion in published data as to the influence of increased height above sea level on the intensity of direct solar radiation. Two indisputable facts are that the intensity is about 1367 W m-2 on a surface normal to the sun’s rays, at the limits of the atmosphere, but only about 1025 W m-2, as a maximum, at the surface of the earth. Thus the intensity is reduced by about 25 per cent when the path length of the sun’s rays is at minimum.

This minimum path-length occurs when the sun is at its zenith or, put another way, when the ‘air-mass’ is unity. The bulk of the earth’s atmosphere is below 3000 m and hence the major part of the reduction in intensity occurs below this level. The reduction in intensity, being clearly bound up with the air mass, must also be dependent on the altitude of the sun, since this affects the path-length of the sun’s rays. (It is approximately true to assume that the path-length is proportional to the cosecant of the angle of altitude of the sun for values of this from 90° to 30°.)

Experimental observations show that for altitude variations from 0 to 3000 m above sea level, the maximum intensity of direct radiation normal to the sun’s rays alters from about 950 W m-2 to about 1170 W m-2. That is to say, an average increase of about 2.5 per cent occurs for each 300 m of increase of height above sea level. It should be noted that as the intensity of direct solar radiation increases with height the intensity of sky radiation falls off. Table 7.2 refers to this.

(4) Variation of the dew point

Although the dew point of the air falls with increasing height above sea level (see equation (2.10) and section 2.13), the effect of this lapse rate is already taken account of in the altitude correction mentioned above. That a variation in dew point has any effect at all stems from the fact that a change in dew point means a change in moisture content and this, in turn, means a change in the absorption capacity of the air-water vapour mixture which constitutes the lower reaches of our atmosphere.

However, dew point varies over the surface of the earth between places of the same height above sea level. Some variation in intensity is, therefore, to be expected on this count. The Carrier Air Conditioning Company (1965) suggests that an increase in the intensity of direct radiation of 7 per cent occurs for each 5 K reduction in dew point below 19.4°C.

(5) Hemisphere

As has been mentioned in section 7.1, the sun is 3 per cent closer to the earth in January than it is in July. This results in an increase of about 7 per cent in the value of the intensity of radiation reaching the upper part of the earth’s atmosphere in January. Hence, calculations carried out for summer in the southern hemisphere should take account of this increase.

EXAMPLE7.il

Calculate the instantaneous transmission of solar radiation through a window recessed 300 mm from the outer surface of the wall in which it is set. Use the following data:

Single 4 mm ordinary clear glazing, facing south-west

TOC o "1-5" h z Latitude 40°N

Outside air temperature (t0) 32°C

Room air temperature (tr) 24°C

Sun time 13.00 h on 23 July

Altitude of location 600 m above sea level

Moderate industrial haze

The window framework is steel and the size of the opening in the wall is 3 m x 3 m. The surrounding ground is covered with grass.

U glass = 5.86 W m~2 K-1 hso glass = 22.7 W nT2 K"1 hsi glass = 7.9 W m~2 K-1

Answer

It is convenient to divide the calculations into a number of steps, in order to illustrate methods.

(a) Calculation of solar altitudes

Declination, d, is 20-y° (see section 7.5) and hour angle, h, is 15° (see section 7.4). Then by equation (7.3):

Sin a = sin 20-y° sin 40° + cos 20° cos 40° cos 15°

= 0.3461 x 0.6428 + 0.9382 x 0.7660 x 0.9659 = 0.9166

Whence a is 66°26′.

The CIBSE table of solar altitude gives a similar result.

(b)

Sin 15°___________ An0 tar, -7r,_ 0.2588 |

Calculation of solar azimuth

Tan z =———————————————— r~ (see equation (7.4))

Sin 40° cos 15° — cos 40° tan 20—°

0.6428 x 0.9659 — 0.7660 x 0.3689 whence z is 37°25′, west of south, or 217° clockwise from north.

(c) Determination of wall-solar azimuth

Figure 7.13 shows that the wall-solar azimuth angle, n, is 45° — 37°25′ = 7°35′.

(d) Calculation of sunlit area By equation (7.10)

X — 300 tan 7°35′ = 39.9 mm

By equation (7.11)

Y = 300 sec 7°35′ tan 66°26′

= 300 x 1.009 x 2.292 55 = 694 mm

Fig. 7.13 Determining the wall-solar azimuth angle for Example 7.11. |

Sunlit area = (3.0 — 0.0399)(3.0 — 0.694)

= 6.83 m2 Total area = 9 m2

(e) Determination of the intensity of direct radiation

By equation (7.9)

I = A exp(-fi/sin 66°26′)

= 1.085 exp(-0.207/0.9166)

= 0.866 kW m"2 This is normal to the sun’s rays at sea level.

Table 7.1 gives a value of 0.905 kW m-2, by interpolation.

(f) Determination of the intensity of sky radiation

By interpolation, Table 7.7 (for the intensity of scattered solar radiation from a clear sky on a vertical surface) gives a value of 59 W m~2 at sea level. The footnote to Table 7.2 quotes a decrease of about 20 per cent, by extrapolation, for a height of 600 m above sea level where the value of sky radiation therefore becomes 47 W irf2.

There should be a correction applied to this figure to take account of the additional scattering resulting from industrial haze. However, data are scanty, the effect is small, and so it is ignored here.

(g) Determination of the effect of haze and height above sea level upon direct radiation

Table 7.2 also quotes an increase of 6 per cent for direct radiation and, by inference, for ground radiation also. A haze correction is 0.9 and so the total correction factor is 1.06 x 0.9 = 0.954.

(h) Calculation of the intensity of ground radiation

From Table 7.7, the intensity of radiation scattered from grass at sea level and for a solar altitude of 66°26′ is 89 W m-2, by interpolation. This intensity is normally incident on the window.

At 600 m the figure would be 1.06 x 89 = 94 W m~2.

(i) Calculation of the intensity of direct radiation normal to the window

By equation (7.6)

/v = 866 cos 66°26′ cos 7°35′

= 343 W m-2, uncorrected for height and haze

Table 7.1 gives 356 W nT2, by interpolation.

(j) Direct radiation transmitted through the window

By equations (7.6) and (7.8)

Cos i = cos a cos n

= cos 66°26′ cos 7°35′ = 0.3963

Whence i = 66°39′ and the transmissivity for this angle of incidence is 0.71, according to Table 7.5. Thus the transmission is 0.71 x 343, or 244 W m-2, uncorrected for haze and height.

(k) Scattered radiation transmitted

Taking an approximate value of 0.79 for the transmissivity of scattered radiation, as recommended in Table 7.5, then, at sea level, the scattered radiation transmitted would be

0.79(59 + 91) = 118 WnT2

At an altitude of 600 m, ignoring the effect of haze, the transmission would be:

0.79(47 + 94) = 111 W nT2

(I) Direct plus scattered radiation transmitted

Through glass in sunlight, the radiation transmitted is due to direct and scattered sources and at sea level it is

= (244+ 118) = 362 W nT2

This value, even when corrected for altitude and haze, is of little use if only part of the window is in sunlight. Scattered radiation is transmitted through the entire window, but direct radiation is transmitted through the sunlit portion only.

Hence, transmitted direct radiation, corrected for haze and height

= 0.954 x 244 x sunlit area = 0.954 x 244 x 6.83 = 1590 W

Transmitted scattered radiation, corrected for height but not for haze

= 111 x total area = 111 x 9 = 999 W

Hence, the transmitted radiation is 1590 + 999 = 2589 W. As is seen later, this does not constitute an immediate load on the air conditioning system.

(m) Transmission due to absorbed solar radiation

Assume a value of 0.06 for the absorption of single glazing with an angle of incidence equal to 66°38′ in the case of both scattered and direct radiation (see Table 7.5).

By equation (7.12)

_ 0.06 x (0.954 x 343 + 47 + 94) + 22.7 x 32 + 7.9 x 24 tg ~ 30.6

= 30.9°C

The convected and radiated heat gain to the interior by transmission and from heat absorbed by the glass is

6.83 x 7.9(30.9 — 24) + (9 — 6.83) x 5.86 x (32 — 24) = 474 W

Total instantaneous solar gain = 2589 + 474 = 3063 W.

Note that the gain because of solar energy absorbed by the glass is quite small. In the figure of 474 W the transmitted gain by virtue of air-to-air temperature difference, is 9 x 5.68 x (32 — 24), or 422 W, and so the gain by absorption is only 52 W, which is 1.7 per cent of the total gain of 3063 W and is evidently almost negligible for ordinary glass.

Posted in Engineering Fifth Edition