In some cases, the treatment of solar radiation is more easily understood at the instant time scale, that is, at the radiance level. Because irradiation over an hour (in Wh/m2) is numerically equal to the mean irradiance during this hour (in W/m2), irradiance values can be, to a certain extent, assimilated to hourly irradiation values. However, since the availability of hourly irradiation data is limited, the problem is how to estimate the hourly irradiation, given the daily irradiation.
To introduce the solution to this problem, it is highly instructive to observe that, in terms of extraterrestrial horizontal radiation, the ratio between irradiance B0(0) and daily irradiation B0d(0) can be theoretically determined from Equations (22.4), (22.14) and (22.15).
where the sunrise angle ms is expressed in degrees and T, the day length, is usually expressed in hours.
From the examination of data from several stations, it has been repeatedly noted [18] that, considering long-term averages of terrestrial radiation, the correspondence between the measured ratio of diffuse irradiance to diffuse daily irradiation, rD = D(0)/Dd(0), and this theoretical expression for extraterrestrial radiation (Equation 22.22) is quite good, while the correspondence between the measured ratio of global irradiance to global daily irradiation, rG = G(0)/Gd(0), and this expression, although not perfect, is quite close, so that a slight modification is required to fit the observed data. The following expressions apply:
0(0) = B0( 0) Dd( 0) Sod(0)
and
where a and b are obtained from the following empirical formulae:
a = 0.409 — 0.5016 x sin(MS + 60)
and
b = 0.6609 + 0.4767 x sin(MS + 60) (22.26)
Note that rD and rG have units of T—1, and that they can be extended to calculate irradiations during short periods centred on the considered instant m. For example, if we wish to evaluate the irradiation over one hour between 10:00 and 11:00 (in solar time), we set m = – 22.5° (the centre time of the considered period is 10:30, i. e. one hour and a half, or 22.5°, before noon) and T = 24 h. If we wish to evaluate the irradiation over one minute, we just have to express T in minutes, that is, we set it to 1440, the number of minutes in a day.
An example can help in the use of these equations: the calculation of the irradiance components at several moments along the 15 April in Portoalegre, Brazil (ф = – 30°), knowing the global daily irradiation, Gd(0) = 3861 Wh/m2. The results are as follows:
ms = -84.51°
—– ms cosoas — sinoAs = 0.8542
180 S S S a = 0.6172 b = 0.4672
rD = 0.0922(cos m + 0.0967) h-1 rG = rD(a + b cos m)
|
M° |
rD [to1] |
rG [Ю1] |
D(0) in [Wm-2] |
G(0) in [Wm-2] |
B(0) in [Wm-2] |
|
Ms |
0 |
0 |
0 |
0 |
0 |
|
±60 |
0.0618 |
0.0529 |
100.94 |
204.25 |
103.31 |
|
±30 |
0.1177 |
0.1211 |
192.24 |
467.58 |
275.34 |
|
0 |
0.1382 |
0.1508 |
225.73 |
582.24 |
356.51 |
Figure 22.12 plots rD and rG versus the solar time, along the day. It is interesting to observe that rG is slightly more sharp-pointed than rD. This is because, due to air mass variations, beam transmittance is higher at noon that at any other moment of the day. Obviously, on integrating, the areas below both, rD and rG must be equal to one. As already mentioned, for this calculation it can be assumed that the irradiation over one hour (in Wh/m2) is numerically equal to the mean irradiance during this hour and also equal to the irradiance at the instant half way through the hour. For example, the global irradiance at noon, G(0) = 580.4 W/m2 can be identified with the hourly irradiation from 11:30 to 12:30, Gh(0) = 580.4 Wh/m2. This assumption does not introduce significant errors and it greatly simplifies the calculations by eliminating the need to evaluate integrals with respect to time, which, otherwise, can be quite tedious.
From these equations, it can be deduced that, on any day of the year and anywhere in the world, 90% of the total global horizontal irradiation is received during a period centred around midday and of length equal to two-thirds the total sunlight day length. Consequently, a stationary receiver tilted to the equator (a = 0) captures all the useful energy in this period. The same need not be true, however, of receivers that track the sun.
It should be mentioned that the ‘average day’ profile given by Equations (22.23) and (22.24) preserves the observed monthly averaged irradiance values, but not the observed distributions of the instantaneous irradiances. Theoretically, this represents a potential source of error on energy estimations, at the extent of PV modules and inverter efficiency depend on the irradiance in a nonlinear fashion. To overcome this problem, some authors [19] consider a weather pattern consisting only of instances of clear sky and instances of overcast conditions with almost zero irradiance. Depending on the frequency of the clear-sky conditions, the average could take any value. This
|
Figure 22.12 Plots of irradiance to daily irradiation ratios, for both diffuse and global radiation, rD and rG, during the day at a latitude ф = – 30° for 15 April |
way, energy estimations for a month are performed on the basis of its ‘clear-sky’ profile and clear – sky index (ratio between the observed average daily irradiation Gdm(0) and the daily irradiation given by the integral of the clear-sky profile). A very simple clear-sky model [20] consists of determining the direct component of the horizontal irradiance by means of Equation (22.13) and the diffuse component as 20% of the direct component. More accurate models [21] require information concerning the turbidity of the sky, which are today available for only some regions. The irradiance profile affects estimations of irradiation over inclined surfaces, inverter efficiency and PV module efficiency and spectral response. Nevertheless, it should be noted that nonlinearities on efficiency-irradiance relations are usually small, so the energy impact of the irradiance profile is also small, typically bellow 2% in terms of daily energy values,
