where t uc is the Coordinated Universal Time (Greenwich Meridian Time), CL the
longitude correction given by the product of −1/15 (60) min per longitude degree
west of Greenwich, and ET is the time equation:
ET ¼ 9:87 sinð2BÞ À 7:53 cosðBÞ À 1:5 sinðBÞ
ð 7:7Þ
where B is given by
B ¼ 360
t j À 81
364
ð7:8Þ
Making the substitutions, the zenith angle w will be 17.34º. This implies that the
solar height is 72.66º or considering that the roof inclination is 21.8º, it means that
the roof at 12 pm is completely exposed to the sun.
The azimuthal angle a is given by Eq. (6.73):
sin a
ð Þ ¼ Àcosdsinh=sinw
ð6:73Þ
Making the necessary substitutions in Eq. (6.73), we get that the azimuthal
angle, representative of the horizontal projection of the solar rays with the true
North, will be equal to 30.79º.
Using Eq. (6.76) in Chap. 6, it becomes possible to calculate the intensity of
instant solar radiation on a horizontal surface outside the atmosphere as
S h ¼ S o 1 þ 0:033 cos 360t j =365
À
Á
Â
Ã
sin b
ð6:77Þ
Replacing the respective values, the intensity S h is 1268.04 W/m
2
.
Next, the calculation of the global direct and diffuse solar radiation on a plane
horizontal to the Earth’s surface (Eq. 6.88) is as follows:
S t ¼ S 0 s
m cos w þ S d 0:271 À 0:294t
m
ð
Þ cos w
ð6:88Þ
where s is the atmospheric transmittance to direct radiation with an arbitrary value
of 0.7, S o the solar constant 1373 Wm
−2 , and m the mass of air given by Eq. (6.82):
m ¼ sec w
ð6:81Þ
which in this case gives 1.05.
The global solar radiation (Eq. 6.88) per unit area (flux density) is 959.75
Wm
−2 . This corresponds to 872.67 Wm
−2 for the direct solar radiation (first term on
the right side) and 87.08 Wm
−2 for diffuse solar radiation (second term on the right
side). The total direct incident radiation on the building (Fig. 7.8A) can now be
calculated. Its value is the product S b A h obtained from Eq. (6.63) between the direct
solar radiation flux per unit area S b , 872.67 Wm
−2 , previously calculated and the
shadowed area on a horizontal surface A h :
258
7 Examples of Applications
longitude correction given by the product of −1/15 (60) min per longitude degree
west of Greenwich, and ET is the time equation:
ET ¼ 9:87 sinð2BÞ À 7:53 cosðBÞ À 1:5 sinðBÞ
ð 7:7Þ
where B is given by
B ¼ 360
t j À 81
364
ð7:8Þ
Making the substitutions, the zenith angle w will be 17.34º. This implies that the
solar height is 72.66º or considering that the roof inclination is 21.8º, it means that
the roof at 12 pm is completely exposed to the sun.
The azimuthal angle a is given by Eq. (6.73):
sin a
ð Þ ¼ Àcosdsinh=sinw
ð6:73Þ
Making the necessary substitutions in Eq. (6.73), we get that the azimuthal
angle, representative of the horizontal projection of the solar rays with the true
North, will be equal to 30.79º.
Using Eq. (6.76) in Chap. 6, it becomes possible to calculate the intensity of
instant solar radiation on a horizontal surface outside the atmosphere as
S h ¼ S o 1 þ 0:033 cos 360t j =365
À
Á
Â
Ã
sin b
ð6:77Þ
Replacing the respective values, the intensity S h is 1268.04 W/m
2
.
Next, the calculation of the global direct and diffuse solar radiation on a plane
horizontal to the Earth’s surface (Eq. 6.88) is as follows:
S t ¼ S 0 s
m cos w þ S d 0:271 À 0:294t
m
ð
Þ cos w
ð6:88Þ
where s is the atmospheric transmittance to direct radiation with an arbitrary value
of 0.7, S o the solar constant 1373 Wm
−2 , and m the mass of air given by Eq. (6.82):
m ¼ sec w
ð6:81Þ
which in this case gives 1.05.
The global solar radiation (Eq. 6.88) per unit area (flux density) is 959.75
Wm
−2 . This corresponds to 872.67 Wm
−2 for the direct solar radiation (first term on
the right side) and 87.08 Wm
−2 for diffuse solar radiation (second term on the right
side). The total direct incident radiation on the building (Fig. 7.8A) can now be
calculated. Its value is the product S b A h obtained from Eq. (6.63) between the direct
solar radiation flux per unit area S b , 872.67 Wm
−2 , previously calculated and the
shadowed area on a horizontal surface A h :
258
7 Examples of Applications