# The intensity of direct radiation on a surface

If the intensity of direct solar radiation incident on a surface normal to the rays of the sun is IW m-2, then the component of this intensity in any direction can be easily calculated. Two simple cases are used to illustrate this, followed by a more general case.

Case 1 The component of direct radiation normal to a horizontal surface (/h), as illustrated in Figure 7.7(a).

If the angle of altitude of the sun is a, then simple trigonometry shows that the component in question is I sin a.

H = I sin a (7.5)

Case 2 The component of direct radiation normal to a vertical surface (/v), as illustrated in Figure 1.1(b).

The situation is a little more complicated since the horizontal component of the sun’s rays, I cos a, has first to be obtained and then in its turn, has to be further resolved in a direction at right angles to a vertical surface which has a particular orientation: that is, the wall-solar azimuthal angle n must be worked out. It then follows that the resolution of / cos a normally to the wall is I cos a cos n.

/v = I cos a cos n (7.6)

EXAMPLE 7.5

If the altitude and azimuth of the sun are 62 ° and 82 j0W of S respectively, calculate the intensity of direct radiation normal to (a) a horizontal surface and (b) a vertical surface facing south-west.

Answer

Denote the intensity of direct radiation normal to the rays of the sun by the symbol /,

L = I sin a |

Vector representing the vertical component, of the intensity |

(b) Fig. 7.7 The intensity of solar radiation on a horizontal and vertical surface. |

HTTTTTTTTTT Horizontal surface (a) |

Vector representing the intensity, /, of the sun’s radiation on a plane normal to its rays T7/777 |

/// |

Having units of W itT2, and denote its components normal to a horizontal and a vertical surface by /h and /v, respectively.

(a)

**4 **= 7 sin a = I sin 62 °

= 0.887/

*[b) *By drawing the diagram shown in Figure 7.8 it is evident that the wall-solar azimuthal angle, n, is 37 j0 and hence

Iv = I cos a cos n = I cos 62 j° cos 37 j°

= 0.366/

Case 3 The component of direct radiation normal to a tilted surface (7§).

Figure 7.9 illustrates the case and shows, in section, a surface that is tilted at an angle 8 to the horizontal. The surface has an orientation which gives it a wall-solar azimuthal angle of n.

If the surface were vertical, then the resolution of I, normal to its surface, would be I cos

Fig. 7.9 The derivation of an expression for the intensity of solar radiation normally incident on a surface tilted at an angle 8 to the horizontal. |

A cos n. The sun’s rays also have a resolution normal to a horizontal surface of I sin a. It follows that the rays normally incident on a tilted surface instead of a vertical one have a vector sum of I sin a and I cos a cos n, in an appropriate direction. The components of I sin a and I cos a cos n in this appropriate direction are I sin a cos 8 and I cos a cos n sin 8, respectively, both normal to the tilted surface. Their sum is the total value of the intensity of radiation normally incident on the tilted surface—the answer required. Hence,

/5 = / sin a cos 8 ± / cos a cos n sin 8 (7.7)

Equation (7.7) degenerates to equation (7.5) for a horizontal surface where 8 is zero, and to equation (7.6) for a vertical surface where 8 is 90°. The significance of the alternative signs is that if the tilted surface faces the sun, a positive sign should be used and vice versa if the surface is tilted away from the sun. Naturally enough, if the surface is tilted so much that it is in shadow, I is zero, since the angle of incidence of the radiation will be 90°. Equation (7.7) can be written in a more general but less informative manner:

/g = I cos i (7.8)

Where i is the angle of incidence of the ray on the surface. This is true of all surfaces, but it must be borne in mind that the angle of incidence is between the incident ray and the normal to the surface. It is not the glancing angle.

All that equation (7.8) does is to state in a succinct form what equation (7.7) does more explicitly. This means that cos i is the same as sin a cos 8 ± cos a cos n sin 8.

EXAMPLE 7.6

Calculate the component of direct solar radiation which is normally incident upon a surface tilted at 30° to the horizontal and which faces south-west, given that the solar altitude and azimuth are 62 j ° and 82 j ° west of south, respectively. The tilted surface is facing the sun.

Answer

For a wall facing south-west and an azimuth of 82 0 west of south, the value of the wall solar azimuthal angle, n, is 31 j°; hence, by equation (7.7),

/5 = / sin 62 ° cos 30° + / cos 62 j 0 cos 37 j 0 sin 30°

= /(0.887 x 0.866 + 0.462 x 0.793 x 0.5)

= 0.95/.

Doing the same calculation by means of the data published in the CIBSE Guide A2 (1999) yields the same answer.

Posted in Engineering Fifth Edition