Estimate the shading fraction of a south-facing window 2 m in height, located in 40° latitude at 10 am and 3 pm on June 15. The overhang is wide enough to neglect the side effects and its length is 1 m, located 0.5 m above the top surface of the window. The window is tilted 15° from vertical and faces due south.

From Example 2.6, on June 15, 6 = 23.35°. The hour angle at 10 am is -30° and at 3 pm is 45°. From the problem data, we have P = 1 m, G = 0.5 m, H = 2 m, ß = 75°, and Zs = 0°. Therefore, from Eqs. (6.55) and (6.56) we have the following.

At 10 am sin (ac) = sin (75) cos (40) cos(23.35)cos(-30)

- cos(75)sin(0)cos(23.35)sin(-30) + sin(75)sin(40)sin(23.35) + cos (75) cos (0) cos (40) sin (23.35)

+ sin (75) cos (0) sin (40) cos (23.35) cos (-30) + sin (75) sin (0) cos (23.35) sin (-30) + cos(75)sin(40)sin(23.35) - sin (75) cos (0) cos (40) sin (23.35) = 0.4240

At 3 pm sin (ac) = sin (75) cos (40) cos (23.35) cos (45)

+ sin (75) sin (40) sin (23.35) + cos(75)cos(0)cos(40)sin(23.35) = 0.6970

+ sin(75) cos(0) sin(40) cos(23.35) cos(45) + sin(75) sin(0) cos(23.35) sin(45) + cos(75)sin(40)sin(23.35) - sin(75) cos(0) cos(40) sin(23.35) = 0.3045

F = PsiniO.) _ G = 1x06970 _ 05 = 0.895, or 89.5% Hcos(9c) H 2 x 0.3045 2

The equation giving the area-average radiation received by the partially shaded window, by assuming that the diffuse and ground-reflected radiation is isotropic, is similar to Eq. (2.97):

where the three terms represent the bean, diffuse, and ground-reflected radiation falling on the surface in question.

The factor Fw in the first term of Eq. (6.57) accounts for the shading of the beam radiation and can be estimated from Eq. (6.53) by finding the average of F for all sunshine hours. The third component of Eq. (6.57) accounts for the ground-reflected radiation and, by ignoring the reflections from the underside of the overhang, is equal to [1 - cos(90)]/2, which is equal to 0.5. The second factor of Eq. (6.57) accounts for the diffuse radiation from the sky, and the view factor of the window, Fw-s, includes the effect of overhang. It should be noted that, for a window with no overhang, the value of Fw-s is equal to [1 + cos(90)]/2, which is equal to 0.5 because half of the sky is hidden from the window surface. The values with an overhang are given in Table 6.1, where e is the relative extension of the overhang from the sides of the window, g is the relative gap between the top of the window and the overhang, w is the relative width of the window, and p is the relative projection of the overhang, obtained by dividing the actual dimensions with the window height (Utzinger and Klein, 1979).

A monthly average value of Fw can be calculated by summing the beam radiation with and without shading over a month:

J GbRbFwdt

GbRb dt

Therefore, the mean monthly and area-average radiation on a shaded vertical window can be obtained by an equation similar to Eq. (2.107):

g |
w |
p = 0.1 |
p = 0.2 |
p = 0.3 |
p = 0.4 |
p = 0.5 |
p = 0.75 |
p = 1.0 |
p = 1.5 |
p = 2.0 |

Values for |
e = 0.00 | |||||||||

0.00 |
1 |
0.46 |
0.42 |
0.40 |
0.37 |
0.35 |
0.32 |
0.30 |
0.28 |
0.27 |

4 |
0.46 |
0.41 |
0.38 |
0.35 |
0.32 |
0.27 |
0.23 |
0.19 |
0.16 | |

25 |
0.45 |
0.41 |
0.37 |
0.34 |
0.31 |
0.25 |
0.21 |
0.15 |
0.12 | |

0.25 |
1 |
0.49 |
0.48 |
0.46 |
0.45 |
0.43 |
0.40 |
0.38 |
0.35 |
0.34 |

4 |
0.49 |
0.48 |
0.45 |
0.43 |
0.40 |
0.35 |
0.31 |
0.26 |
0.23 | |

25 |
0.49 |
0.47 |
0.45 |
0.42 |
0.39 |
0.34 |
0.29 |
0.22 |
0.18 | |

0.50 |
1 |
0.50 |
0.49 |
0.49 |
0.48 |
0.47 |
0.44 |
0.42 |
0.40 |
0.38 |

4 |
0.50 |
0.49 |
0.48 |
0.46 |
0.45 |
0.41 |
0.37 |
0.31 |
0.28 | |

25 |
0.50 |
0.49 |
0.47 |
0.46 |
0.44 |
0.39 |
0.35 |
0.27 |
0.23 | |

1.00 |
1 |
0.50 |
0.50 |
0.50 |
0.49 |
0.49 |
0.48 |
0.47 |
0.45 |
0.43 |

4 |
0.50 |
0.50 |
0.49 |
0.49 |
0.48 |
0.46 |
0.43 |
0.39 |
0.35 | |

25 |
0.50 |
0.50 |
0.49 |
0.48 |
0.47 |
0.44 |
0.41 |
0.35 |
0.30 | |

Values for |
e = 0.30 | |||||||||

0.00 |
1 |
0.46 |
0.41 |
0.38 |
0.33 |
0.33 |
0.28 |
0.25 |
0.22 |
0.20 |

4 |
0.46 |
0.41 |
0.37 |
0.34 |
0.31 |
0.26 |
0.22 |
0.17 |
0.15 | |

25 |
0.45 |
0.41 |
0.37 |
0.34 |
0.31 |
0.25 |
0.21 |
0.15 |
0.12 | |

0.25 |
1 |
0.49 |
0.48 |
0.46 |
0.43 |
0.41 |
0.37 |
0.34 |
0.30 |
0.28 |

4 |
0.49 |
0.47 |
0.45 |
0.42 |
0.40 |
0.34 |
0.30 |
0.24 |
0.21 | |

25 |
0.49 |
0.47 |
0.45 |
0.42 |
0.39 |
0.33 |
0.29 |
0.22 |
0.18 | |

0.50 |
1 |
0.50 |
0.49 |
0.48 |
0.47 |
0.45 |
0.42 |
0.39 |
0.35 |
0.33 |

4 |
0.50 |
0.49 |
0.47 |
0.46 |
0.44 |
0.39 |
0.34 |
0.27 |
0.26 | |

25 |
0.50 |
0.49 |
0.47 |
0.46 |
0.44 |
0.39 |
0.34 |
0.27 |
0.22 | |

1.00 |
1 |
0.50 |
0.50 |
0.49 |
0.49 |
0.48 |
0.47 |
0.45 |
0.42 |
0.40 |

4 |
0.50 |
0.50 |
0.49 |
0.48 |
0.48 |
0.45 |
0.43 |
0.38 |
0.34 | |

25 |
0.50 |
0.50 |
0.49 |
0.48 |
0.47 |
0.44 |
0.41 |
0.35 |
0.30 |

Methods to estimate HD/H and RB are described in Section 2.3.8 in Chapter 2. An easy way to estimate Fw is by using Eq. (6.53) and finding the average of F for all sunshine hours for the recommended average day for each month, shown in Table 2.1.

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