18.3
d′ = l cos θ M = 857 mm.
Thus, we can mount four rows behind each other, because
3d + d′ = 3 × 2430 + 857 = 8147 mm,
which is less than 10 m. In one row fit 6 modules because 6 × 1,650 = 9,900 mm. Thus, we can place 24 modules.
Direct and diffuse irradiance
As sunlight traverses the atmosphere, it is partially scattered, leading to attenuation of the
direct beam component. On the other hand, the scattered light will also partially arrive at
on the Earth’s surface as diffuse light. For PV applications it is important to be able to
estimate the strength of the direct and diffuse components.
First, we discuss a simple model that allows us to estimate the irradiance on a
cloudless sky independent of the air mass and hence the altitude of the Sun. As we have
seen in Section 5.5, the air mass is defined as
where the angle between the Sun and the zenith θ is connected to the solar altitude via θ =
90° − a S . This equation, however, does not take the curvature of the Earth into account. If
the curvature is taken into account, we find [138]
To estimate the direct normal irradiance at a certain solar altitude a S and altitude of
the observer h, we can use the following empirical equation [139]
with the constant c = 0.14. The solar constant is given as
Wm
-2 . Even during
clear sky conditions the diffuse irradiance is about 10% of the direct irradiance. Thus the
global irradiance on a module perpendicular to the Sun can be approximated as [140]
For a high diffusion percentage this approach no longer works very well. A more accurate
model was developed in the framework of the European Solar Radiation Atlas [141]. In
that model the direct irradiance for clear sky is given by
I 0 is the solar constant that takes a value of 1,361 Wm
-2 . The factor ε corrects for
deviations of the Sun-Earth distance from its mean value. a S is the solar altitude angle.
d′ = l cos θ M = 857 mm.
Thus, we can mount four rows behind each other, because
3d + d′ = 3 × 2430 + 857 = 8147 mm,
which is less than 10 m. In one row fit 6 modules because 6 × 1,650 = 9,900 mm. Thus, we can place 24 modules.
Direct and diffuse irradiance
As sunlight traverses the atmosphere, it is partially scattered, leading to attenuation of the
direct beam component. On the other hand, the scattered light will also partially arrive at
on the Earth’s surface as diffuse light. For PV applications it is important to be able to
estimate the strength of the direct and diffuse components.
First, we discuss a simple model that allows us to estimate the irradiance on a
cloudless sky independent of the air mass and hence the altitude of the Sun. As we have
seen in Section 5.5, the air mass is defined as
where the angle between the Sun and the zenith θ is connected to the solar altitude via θ =
90° − a S . This equation, however, does not take the curvature of the Earth into account. If
the curvature is taken into account, we find [138]
To estimate the direct normal irradiance at a certain solar altitude a S and altitude of
the observer h, we can use the following empirical equation [139]
with the constant c = 0.14. The solar constant is given as
Wm
-2 . Even during
clear sky conditions the diffuse irradiance is about 10% of the direct irradiance. Thus the
global irradiance on a module perpendicular to the Sun can be approximated as [140]
For a high diffusion percentage this approach no longer works very well. A more accurate
model was developed in the framework of the European Solar Radiation Atlas [141]. In
that model the direct irradiance for clear sky is given by
I 0 is the solar constant that takes a value of 1,361 Wm
-2 . The factor ε corrects for
deviations of the Sun-Earth distance from its mean value. a S is the solar altitude angle.