Efficiency comparison between tracking and optimally fixed flat solar collectors

We investigate the optimal orientation for a fixed flat plate solar collector using the clear sky model. The ground reflection component of irradiation that hits the collector’s surface is ignored due to its relatively small magnitude when compared to the direct beam and sky diffusive components. Analytical calculations demonstrate that regardless of the collector’s latitude, the most effective azimuthal angle, \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\gamma ^*$$\end{document}γ∗, is 0, which generally corresponds to a North–South direction. However, the optimal tilt angle, \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\beta ^*$$\end{document}β∗, is dependent on both the Day of Year (DoY) and the collector’s local latitude. For latitudes typical of mid-altitude climate zones, we can calculate the optimal tilt angle and the maximum energy that the collector can harvest during each DoY. We compare the maximum daily received energy—which is the sum of the direct beam and sky diffusive energies—associated with this optimal orientation to their corresponding values when the flat plate tracks the Sun. The relative increase in total energy due to Sun tracking depends critically on the DoY, with a minimum value of about \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$17\%$$\end{document}17% in early winter and a maximum value of \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$40\%$$\end{document}40% over a large interval.


Some astronomy
To describe the Sun-Earth geometry, one needs a system of coordinates.There are two main pictures: heliocentric and geocentric.In the heliocentric picture, the Sun (at the origin) is placed at one of the foci of an ellipse with a small eccentricity around 0.0167.The Earth orbits around the Sun in an elliptical path on a plane called the ecliptic.The plane perpendicular to the Earth axis (which connects the North pole to the South one at the Earth's center) is the Earth equatorial plane.It is inclined to the ecliptic plane by an obliquity angle 23.45 • .In the geocentric system of coordinates, the Earth is at the origin.There are two choices for the x − y plane: equatorial and horizontal which will be briefly explained.
Horizontal picture.In the horizontal perspective, the system of coordinates is established with the observer as the origin, the observer's horizon as the fundamental x − y plane, and the z-axis along the zenith, i.e., the observer's overhead direction towards the sky.The Sun's angular position is described by two angles of elevation, or altitude, α s , and azimuth γ s .Figure 1 provides an illustration.The zenith angle, θ z , and the solar altitude angle, α s , are complementary.So, θ z = π 2 − α s .This angle represents the direct beam irradiation angle.The unit vector σ specifies the line connecting the observer to the Sun, while the unit vector ν specifies the North pole direction, i.e., the line connecting the Earth's center to its North pole.
Equatorial picture.In the equatorial perspective, the origin is placed at the Earth's center.In this picture the fundamental plane is the equatorial plane that passes through the terrestrial equator.The Sun's angular position is specified by two angles: declination, δ , and hour angle, ω .The angle between the equatorial plane and the line connecting the Sun to the origin (Earth) is the declination angle, shown by δ .Alternatively, the declination angle, δ , is the Sun's altitude with respect to the equatorial plane.To specify the hour angle, ω , we first need to define a local meridian.Earth is at the origin of the Celestial sphere.The local celestial meridian is the circle on the celestial sphere.It is perpendicular to both the horizontal plane and the equatorial plane.The zenith, nadir, North Celestial Pole, and South Celestial Pole are located on the celestial meridian.The hour angle, ω , is defined as follows: first, project the line connecting the origin to the Sun onto the equatorial plane.The hour angle ω is

Formulation and methodology: optimal orientation of a flat plate receiver
It's clear that the maximum amount of energy can be received on the collector's aperture if it tracks the Sun's movement, but as we have noted, such a tracking mechanism can be expensive.When we refer to the optimal configuration of a solar collector, we are talking about specifying the collector's fixed orientation, which includes the two angles of tilt and azimuth.The objective is to determine the orientation that maximizes the total amount of irradiation that impinges on the collector's surface during a given time period.The time period that is spent to measure the irradiation time may vary from a single day to a month, a season or even a year depending on our desired application.For the purpose of determining the optimal orientation for a solar collector, we will consider the shortest period, which is one day.With daily data for the optimal orientation, the procedure for longer periods can be easily extrapolated.For our purposes, we will focus on the simplest collector geometry, which is a flat plate collector with a unit area.The flat plate collector is located at a local latitude of ϕ , and its orientation is determined by the direction of its unit normal vector, n .In the spherical coordinates system of an observer, where the x − y plane is the local horizontal plane and the k direction is aligned with the zenith, this unit normal vector can be specified by two angles: the tilt angle, β , and the azimuthal angle, γ .The unit vector can be decomposed as follows: (1) σ =ν sin δ + e w cos δ sin ω + u cos δ cos ω, (2) σ =k sin α s + e s cos α s cos γ s + e w cos α s sin γ s , , then positive (negative) values for β indicate that the panel is oriented towards the South (North).The unit vector n can be expanded in terms of ν and the two other perpendicular unit vectors of the equatorial plane in a similar manner.The incidence angle θ is defined as the angle between the normal to the plate, n , and the direction of the Sun, which is the line connecting the observer to the Sun, denoted by σ .In the equatorial plane n can be decomposed as: The angle between n and ν is denoted by η , while the angle between the projection of n onto the equatorial plane and the unit vector u within that plane is denoted by ζ .Please refer to Fig. 4 for an illustration of these angles.
By utilizing Eqs. ( 1), ( 2), (6), and (7), we arrive at Equations ( 8) and ( 9) provide cos θ in terms of the Sun's positional angles ( δ and ω ) in the geocentric equatorial system of coordinates, the observer's latitude ϕ , and the flat plate's orientation angles, β and γ , or equivalently, (6) n = k cos β + e s sin β cos γ + e w sin β sin γ .Our objective is to determine the total amount of radiation energy that the flat plate receives on the nth day of the year, where n = 1 corresponds to the first of January in the Julian calendar.The total instantaneous irradia- tion, G, that impinges on a surface is composed of three components: beam direct radiation, G b , sky diffuse radiation, G d , and ground diffuse reflection, G r .The beam direct radiation, G b , typically provides the largest contribution to the total irradiation.Since the ground reflection component is small compared to G b and G d , it will be neglected in this paper.To begin, we will focus on the energy, E b , received by the flat collector due to the beam direct irradiation, G b .
Contribution of beam direct radiation to received energy.The total direct irradiation energy during a day (from sunrise to sunset) received on a flat plate of unit area is: where the variables t r and t s represent the local times of sunrise and sunset, respectively.G bn is the magnitude of the direct normal beam, and G bn = G bn (−σ ) represents the direct normal beam vector.It should be noted that t r and t s are functions of n and ϕ , but for brevity, we do not explicitly write them.The variable is the Heaviside step function, which ensures that the surface receives irradiation when the following condition holds To evaluate the integral in Eq. ( 12), we need to specify the dependence of the Sun's geocentric coordinate angles δ and ω on the day of the year (DoY) and local time.The declination angle has a weak dependence on local time, so we neglect it in this paper unless stated otherwise.The dependence of the declination angle δ (in radians) on the day of the year (n) is given by 43,44 : In contrast to the declination angle, the hour angle ω is solely a function of time, specifically the hour of the day (HoD).We use solar time, which is based on the apparent angular motion of the Sun across the sky.Since the Earth rotates by 15 • per hour around its axis, the relationship between solar time and hour angle in terms of degrees (radians) per second can be expressed as: Defining ω = 0 at solar noon ( h = 12 : 00 h ) the hour angle ω can be determined for any solar time hour.The last step would be to calculate the local times t r and t s (for a given DoY, n, and latitude ϕ ) in terms of solar time t sol .For this purpose, let us first write the equation relating local time (standard time) t std and solar time t sol : with EoT (Equation of Time): 365 44 , L loc is the local longitude, and L r = (LCT − GMT) × 12.5 • /hour is the reference longitude.
LCT is the local civil time and GMT is Greenwich Mean Time.For example, Tehran civil time is GMT+3.5 hours hence for Tehran we have: L r = 3.5 × 15 • = 52.5 • .Principally we change the integral variable from standard time t into solar time t sol in Eq. ( 12).However, we need a second change of variable from solar time t sol to hour angle variable ω in the integrand.It turns out: where ω r is the Sun's hour angle at sunrise and ω s is the Sun's hour angle at sunset.These are obtained from (3) by setting α s = 0 .This equation ( cos ω = − tan ϕ tan δ ) has two solutions ω r and ω s = −ω r correspondingly.
The hour angle dependence of the direct normal beam, G bn (n, ω) , makes the integral difficult to evaluate.In the (10) n =ν cos η + u sin η, (11)  =k cos β + e s sin β.In the subscript of G on , note that "o ′′ refers to "outside" and "n ′′ refers to "normal".Due to extinction processes such as Rayleigh or Mie scattering, or absorption, the amount of irradiance that reaches the Earth's surface is less than the amount outside of the atmosphere.Taking into account all of these attenuation effects, the amount of irradiance at ground level, denoted by G bn , can be approximated using the following formula: Here, τ b represents the effective atmospheric transmission coefficient of the direct beam.Various models, each with their own set of assumptions and parameters, have been proposed to estimated the amount of τ b .Each model has its own advantages and limitations.In this work our focus is on a clear sky condition where there is no cloud in the sky and the atmosphere over the studied location is free of pollutants.Under this assumption, a wide range of clear sky models exists in the literature.For simplicity, we assume a clear sky model proposed in 46 , in which the sky is cloudless, clear (visible up to 23 km), and pollution-free.According to Hottel's model, the effective atmospheric optical transmission coefficient τ b is: The constants a 0 , a 1 and k are altitude and climate type dependent.See chapter two of 1 for further details.) and finally into Eq. ( 18) we arrive at: The integral given by Eq. ( 22) must be numerically evaluated.The variable of integration, the hour angle ω , var- ies from the hour angle at sunrise, ω r , to the hour angle at sunset, ω s .Note that the hour angle ω is measured in radians.We will compute the integral using the Simpson's rule.To proceed, we should specify the integral limits ω r and ω s .For this purpose, we take the city Tehran with latitude ϕ = 35.69Assuming a negative angle, we have ω r = −ω s = −90 • .To evaluate the integrals, we need to specify the numeri- cal values of the transmission coefficient parameters a 0 , a 1 , and k.These values depend on the climate type, and are given by 1 : where Here, A represents the observer's altitude in kilometers, and the correction factors r 0 , r 1 , and r k are dependent on the climate.For Tehran, which is a mid-altitude city with an altitude of A = 1.2 km , the climate-dependent correction factors are r 0 = 0.97 , r 1 = 0.99 , and r k = 1.02 .Taking these factors into account, the transmission coefficient parameters become: (28) a 0 = 0.228; a 1 = 0.666; k = 0.308.
Figure 5 shows the beam irradiation E b on a surface with unit area as a function of the tilt angle β , for various plane azimuthal angles γ , in Tehran on the day of Nowruz.
As shown in Fig. 5 for δ = 0 (corresponding to the day n = 81 ), the beam irradiation energy received by a flat plate is maximized at the optimal angles γ * = 0 and β * = ϕ .Let us prove this analytically.To obtain the beam irradiation energy received by a flat plate with arbitrary orientation β and γ (or η and ζ ), one needs to evaluate the integral: Note that sin η > 0 and it is independent of ω .First, let us approximate the effective atmospheric optical trans- mission coefficient τ b to be a constant.Then we should maximize the following integral Here we have used Eq. ( 9).The Sun's path is symmetric with respect to the plane normal to the equatorial plane passing through the zenith.Due to this symmetry, it is evident that ζ = 0 corresponds to the extremum value of the received energy, and hence J(ζ ) is maximized.For any positive solution of ζ that leads to an optimal value for E, there should exist a negative solution as well.To complete the proof, let us show this directly.We may take 0 ≤ ζ ≤ π 2 without any loss of generality.The same argument applies if we take Note that we have used the identity: x�(x) =   The integrand in the right-hand-side of ( 38) is the multiplication of two functions; an even function with respect to ω , τ b (ϕ, ω) , and an odd function with respect to ω .Then the integral vanishes.Therefore, ζ * = 0 and η * = π 2 (or equivalently γ * = 0 and β * = ϕ ) maximizes the received energy by the flat plate.For a general day of the year (DoY) represented by n, it can be shown, by a symmetry argument, that γ * = 0 remains the optimal azimuth angle, but β * depends crucially on n. Figure 6 shows the dependence of E b on β for different values of n.
Figure 7 shows the daily beam direct irradiation energy (in MJ) received by a flat plate of unit area as a function of its tilt angle β for all days of the year n = 1, • • • , 365 .Each graph in (7) represents the daily beam irradiation energy versus β for a specific day of the year.The daily received energy, takes its maximum value at a day-dependent tilt angle β * (n) .Figure 8 shows the maximum beam irradiation energy (MJ) received by a flat plate of unit area located at Tehran with latitude ϕ = 35.69• N , the plate's azimuth angle is set to zero. Figure 9 shows the optimal tilt angle β * versus DoY, n.
Although we have presented data for a specific location (Tehran, Iran) in the Northern hemisphere, the qualitative behavior of the results remains the same for other locations with similar climate types.

Sky diffuse contribution. Now we investigate the contribution of the sky diffusive irradiation component
E d to the energy received by a tilted flat collector.There are different models that describe the directionality of diffusive irradiation.Here, we adopt the simpler assumption that the sky diffusive part of the solar irradiation is isotropic.The unit vector e r , which represents the direction of the received diffusive radiation, can be specified by two angles: the polar angle ϑ and the azimuth angle φ in the local observer's spherical system of coordinates.We have: See Fig. 10 for illustration.As the symmetry argument still holds true in this case, we set γ * = 0 .Having the unit vector n in Eq. ( 6) we find:  The contribution E d of the sky diffusive component to the received irradiated energy turns out to be where G d (n, ω) is the sky diffusive irradiation on a horizontal surface of unit area and S is the angular integration region satisfying e r • n ≥ 0 .It turns out, Diffuse irradiation appears to be isotropic as there is no preferred direction when we look at the sky.Many models have been proposed to study the contribution of diffuse irradiation.Interested readers may find a review on many of them in 47 .Here we adopt the version of the clear sky model proposed by Liu and Jordan in 48 .According to their model the instantaneous sky isotropic diffusive irradiation G d on a horizontal surface with unit area is given as follows: where τ d = 0.271 − 0.294τ b is the atmospheric transmission coefficient of the diffusive irradiation.The instan- taneous total diffuse irradiation on a tilted surface of unit area is given by: (40) e r • n = sin ϑ sin(β + φ).The daily energy received by a tilted flat plate due to the sky diffusive irradiation becomes: Substituting G dT (n, ω) from ( 44), the integral in (45) becomes: This integral can be numerically evaluated for a given tilt angle β and DoY.Our code is able to find the optimal value β * for the total received energy E tot = E b + E d .Figure 11 shows the maximal daily total energy E tot and beam component E b impinged on a flat plate with unit area.As it is seen the diffusive irradiation enhances the total received energy Figure 12 shows the daily optimum tilt angle β * both for the direct E b and the total E tot energies versus n.As you see, the optimal values are very close to each other.

Comparison with a tracking flat plate
It is highly desirable to determine the amount of radiation energy that a flat collector would receive if it tracks the Sun.As the plate tracks the beam of the Sun, it maximizes its solar exposure by orienting perpendicularly to the beam.To answer this question, the received energy is divided into two parts: direct beam and diffuse sky radiation.Evaluating the contribution of direct beam energy is relatively straightforward.From a mathematical perspective, it is necessary to maintain the incidence angle θ at zero, which means that the unit normal vector n is always along the solar vector σ throughout the day.By equating Eqs. ( 1) and ( 6), we obtain β = π 2 − α s = θ z and γ = γ s .For example, for an observer at the equator ( ϕ = 0 ) and on n = 81 , where δ = 0 , we have according  to (3) β = ω .The total amount of direct beam energy E trc b a tracking flat plate receiver can gain in the nth DoY can be obtained by setting θ = 0 (cos θ = 1) in integral (22).It turns out: The integral in (47) can numerically be computed.Next, the sky diffusive contribution is investigated.For a tracking plate, we can consider it as a fixed plate with an instantaneous tilt angle β with cos β = k • n = k • σ = sin α s .Replacing cos β with sin α s in Eq. ( 44) the instantaneous sky diffusive irradiation of a tracking flat plate becomes: The sky diffusive contribution to the daily received energy of a tracking plate can be obtained via the following integration: Here, we have used the relationship sin α s = cos θ z .Figure 13 shows the daily dependence of the sky diffuse irradiance E d , beam irradiance E b , and the total received energy E tot for a flat plate tracker of unit area located in Tehran, Iran, with a latitude of ϕ = 35.69• N.
As expected, a tracking plate is able to capture more solar energy than a fixed plate with optimal orientation.Figure 14 illustrates the daily variation in the total energy received by both fixed and tracking plates, as well as the percentage increase in energy due to the tracking.
As you can see, the relative amount of the total energy increase percentage due to the Sun tracking critically depends on DoY.The minimum value of the relative energy increase percentage is about 17% in early winter.It raises around 40% in a relatively large interval starting around mid-May.

Comparison to existing results in the literature
Most of the existing results are obtained for particular locations with different weather conditions.As far as we know, no similar results have been found for the city of Tehran.While some investigations have been conducted for cities in south of Iran near the Persian Gulf, none of them have taken clear sky conditions into account.The only papers that have taken the clear sky assumption into account are 14 , which investigated the city of Assiut in Egypt with latitude ϕ = 27.82• N , and 21 , which calculated the optimal tilt angles for representative days of each month at various latitudes on a daily basis.The qualitative dependence of both the daily-averaged optimal tilt angle and the maximal received energy at the daily optimal angle on the day of the year in Tehran and Assiut are similar to each other.This can be seen by comparing Figs.11 and 12 with Fig. 3 of reference 14 .Furthermore, our results for the optimal tilt angle are in good agreement with those presented in 21 for the latitude ϕ = 35 • N , which is relatively close to the latitude of Tehran.To shed more light on the problem, we compared our findings with (47) existing investigations that have taken into account weather conditions.Most of these papers have reported monthly averages at specific locations.To account for weather conditions, including cloudiness and pollutants, a clearness index K is typically introduced.This index represents the ratio of the monthly-averaged total energy received by a horizontal surface (direct, diffusive, and ground reflection) to the energy that a horizontal surface would receive outside of the atmosphere.It is defined as , where H is the monthly-averaged received energy and H o is the monthly average daily energy received outside the atmosphere.The monthly averaged total radiation energy received by a tilted surface of unit area on Earth is given by H T = RH , where R < 1 is a coef- ficient that can be estimated by considering the beam, diffuse, and reflected components of the radiation incidence on the tilted surface individually.Assuming that the diffuse and reflected radiation are isotropic, Liu and Jordan 48 proposed that R can be expressed as: Here, H d is the average diffusive daily energy that a horizontal surface of unit area receives, ρ g is the ground albedo reflection coefficient, and β is the surface tilt angle.Once R b and H d are specified, the monthly average total energy that a tilted surface receives can be calculated.There are various models available for determining H d .Usually, people express H d as a polynomial of the clearness index K .Readers can refer to 27 and 47 for various models of H d and the exact expression of R b .The studies that are closest to the latitude of Tehran were conducted in 49,50 , which investigated eight cities in Turkey.Among them, Adana with a latitude of ϕ = 36.59• N has the closest latitude to Tehran.For example, on March 16, the optimal tilt angles in Tehran (clear sky assumption) and Adana (realistic weather condition) are 33 • and 36 • , respectively.Qualitatively speaking, the daily dependence of the optimal tilt angle is similar in both cities, despite the fact that the total energy is larger in Tehran than in Adana.In another study conducted for higher latitudes in the Northern hemisphere, the optimal tilt angle for Nottingham, England was found to be around 50 • in mid-March.In 21 , the optimal tilt angle for a latitude of 35 • was found to be 38 • in mid-March.Another study conducted for Abu Dhabi, which has a latitude lower than Tehran, gives the average optimal tilt angle for March at 25 • .Naively speaking, the optimal tilt angle tends to decrease with decreasing latitude.

Summary and conclusion
To summarize, we have analytically calculated the optimal tilt and azimuth angles for a fixed flat-type collector at a given latitude.The contribution of the ground-reflected solar radiation is neglected.We have separately evaluated the maximal energy received by a flat plate of unit area associated with the direct beam and sky diffuse components of irradiation.In a geocentric system of coordinates, it has been analytically shown that the optimal azimuthal angle of a collector is γ * = 0 , based on the Sun's symmetrical motion with respect to the equatorial plane.However, the optimal tilt angle β * crucially depends on the day of the year (n) and the local latitude.Analytical analysis has been performed for a mid-altitude climate location in Tehran, Iran.The maximal daily received energies, namely direct beam and sky diffuse, associated with this optimal orientation are compared to their corresponding values when the flat plate tracks the Sun.The relative amount of the total energy increase due to Sun tracking drastically depends on the day of the year.In early winter, the increase percentage is minimum and around 17% .In a relatively large interval starting from mid-May, the increase percentage is large, up to 40% .Although our results are associated with a particular location (Tehran) with a mid-altitude climate, the qualitative behavior for other latitudes remains the same.We have not found such a detailed comparison elsewhere in the literature.

Figure 1 .
Figure 1.Sun's angular position is given by two angles elevation α s and azimuth γ s with the observer's horizon as the fundamental ( x − y ) plane.

Figure 2 .
Figure 2. Equatorial geocentric system of coordinates: Sun angular position is characterised by two angles declination δ and hour ω.

( 9 )Figure 3 .Figure 4 .
Figure 3. Specification of a flat plate orientation by two angles: tilt angle, β , and azimuth angle, γ , in the local observer's system of coordinates, specified by the horizon plane and the zenith.

Figure 5 .
Figure 5. Direct beam irradiation energy E b (MJ) received on n = 81 DoY by a flat plate of unit area versus its tilt angle β for various values of azimuth angle γ .The plate is located at Tehran with latitude ϕ = 35.69• N.

Figure 6 .
Figure 6.Beam direct irradiation energy (MJ) received by a flat plate of unit area on n-th DoY versus its tilt angle β for various values of δ .The plate is located at Tehran with latitude ϕ = 35.69• N .The plate's azimuth angle is set to γ = 0.

Figure 7 .Figure 8 .
Figure 7. Daily beam direct irradiation energy (MJ) received by a flat plate of unit area as a function of its tilt angle β (in radians).The lower (upper) curve corresponds to day of the year n = 354 ( n = 171 ), representing the winter solstice and summer solstice, respectively.The plate is located in Tehran, Iran, with latitude ϕ = 35.69• N , and its azimuth angle is set to γ = 0.

Figure 11 .
Figure 11.Daily maximum total irradiated energy E tot and beam irradiated energy E b (MJ) received by a flat plate of unit area versus n.The plate is located at Tehran with latitude ϕ = 35.69• N .The plate's azimuth angle is set to the optimal value γ * = 0.

Figure 12 .
Figure 12.Plot of the optimum daily tilt angle β * versus n for both E b and E tot .The plate is located at Tehran with latitude ϕ = 35.69• N .The plate's azimuth angle is set to γ = 0.

Figure 13 .
Figure 13.Dependence of separate sky diffusive E d , beam E b and total E tot on n for a tracker flat plate of unit area.The plate is located at Tehran, Iran with latitude ϕ = 35.69• N.

Figure 14 .
Figure 14.Daily dependence of the total energies, fix and tracking (left), and the relative energy increase percentage amount due to the tracking (right).The plate is located at Tehran, Iran with latitude ϕ = 35.69• N.
and ζ .If we consider the special case of γ = 0 (or equivalently, ζ = 0 ), the calculations become simpler.The vector n is in the plane constructed by u and ν ( k and e s ).Therefore, we can express n as follows: η following subsection, we will briefly review this dependence.The solar energy that reaches the Earth is the electromagnetic energy emitted by the Sun which to a good extent can be approximated to be a black body with surface temperature 5777 K .As the light travels a long distance between the Sun and Earth (average distance 1.496 × 10 11 m ) this flux of energy can be assumed to reach the outer region of the Earth's atmosphere in the form of a plane wave.This radiative flux, the energy per unit time received on a surface of unit area perpendicular to the propagation direction, is named solar constant and is denoted by G sc .It can be easily verified that the solar constant has a value G sc = 1367 W m 2 1,45 at the mean Sun-Earth distance.Due to the Earth's orbit eccentricity around the Sun, the solar irradiation outside the Earth's atmosphere (extraterrestrial irradiation) G on has a dependence on DoY.It can be approximated in the following way: ω)]dω.Vol:.(1234567890)Scientific Reports | (2023) 13:12712 | https://doi.org/10.1038/s41598-023-39892-ywww.nature.com/scientificreports/