With renewable energy comes a source intrinsically stochastic hence the group set out to develop a model adaptable to this.

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How does a PV Panel works?

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PV panels are mostly made out of monocrystalline silicon which is a

semiconducting material.

The process of current production is based on a p-n junction. When

light falls on the cell, photons create electron-hole pairs on both 

sides of the p-n junction.

After some time a negative charge develops in the emitter and a 

positive charge in the base.

If a circuit is connected to the emitter and base a current flows.

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How is the PV incorporated into the H2ool?

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We have developed a full trigonometric PV model where a user can input irradiance data specific to his/her location and the model will generate the annual hourly energy output of a specified PV area.

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The following paragraphs will explain in technical detail how the trigonometric model has been developed and works.

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First of all it is important to know where the system would be located. Depending on the

location (Figure 2) and the local time the irradiance data, solar position and solar angles differ.

Therefore the first inputs are: 

• Location

• Latitude and Longitude

• Local time Zone

The model is based on an hourly overview of a full year.

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There are two different types of solar radiation possible, direct or diffuse solar radiation,

which can be seen in the pictures below.

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Since the power output of a PV panel is dependent on the type and the amount of direct or indirect radiation the second part of the required input data is​

• Direct normal solar radiation (W/m^2)​

• Diffuse horizontal solar radiation (W/m^2) 

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Depending on the daytime (Figure 5) and month (Figure 6) the solar radiation falls in different angles on the earth. Therefore, the trigonometric model takes first of all the day in the year into account and calculates the declination, d (deg) and the equation of time, E:

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d = 23.45 * sin (280.1 + 0.9863 Y)

E = 9.87 sin(1.978Y - 160.22) - 7.53 cos(0.989Y - 80.11) - 1.5 sin(0.989Y - 80.11)

 

where Y = day number of the year

(e.g. January 1 =1, December 31 = 365)

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Using the equation of time the local solar time, ts can be calculated. "This is a time that relates to the apparent angular motion of the sun across the sky vault, with solar noon corresponding to the point in time at which the sun traverses the meridian of the observer." (Clarke and Kelly, 2017).

ts = (tm ± time zone) ± LG/15 + E/6

 

where tm = Local clock time

LG= longitude

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Knowing the these parameters allows the calculation first of the solar altitude(deg), then solar azimuth (deg) and finally the incidence angle (s. Figure 7) can be calculated with:

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solar altitude:

 

βs = sin-1 [cos LT cos d cos θh + sin LT sin d ],

where LT  = latitude

 θh is hour angle, 

θh = 15 (12 –  Solar local time)

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solar azimuth: 

 

αs = sin-1 [ cos d sin θh / cos βs ]

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incidence angle of direct beam:

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iβ = cos-1[ sin βs cos (90-βf ) + cos βs cos ω sin (90-βf )],

where ω = surface-solar azimuth 

βf = surface inclination angle

 

surface-solar azimuth (deg) = |αs - αf |

where αs is the solar azimuth 

and αf is surface azimuth

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Assumption: The surface azimuth is 180 deg.

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Inclined surface irradiation

In order to calculate the total radiation incident on an exposed surface of inclination βf and azimuth αf is consistent of three different radiation types which are:

• direct,

• surroundings-reflected and

• sky diffuse. 

Assuming an isotropic sky, the unknown solar quantities are calculated using the angles carried out before.

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To calculate the direct radiation on a surface of inclination (W/m^2) on the PV panel the following formula has been used:

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Idβ = Idh cos iβ /sin βs,

where Idh is the direct horizontal intensity (W/m^2 )

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Then the sky diffuse radiation incident on a surface of inclination (W/m^2) is calculated using:

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Isβ =Ifh{0.5[1+cos(90-βf)]} (isotropic sky)

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Finally also the ground reflected radiation incident on a surface of inclination, Irβ (W/m^2) is taken into account with:

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Irβ = 0.5 [1 - cos (90 - βf)]( Idh + Ifh)rg

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In the end all solar irradiance types are summed up in order to evaluate the total solar irradiance power, ITOT, on the PV panel (W/m^2). With that information the PV panel power output can be determined, using equation:

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Power output = PSTC  x (ITOT/1000) x A

where A = area,

 

PSTC is the power output at Stantard Test Conditions (1000W/m^2 & 25°C). This model assumes a constant temperature but this can be altered. 

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In order to find the day with the average daily PV output, the hourly PV power is summed up over the day.

The average daily PV output can then be matched to the power required by the system. That results into the required PV area given in m^2 and peak power output in W.

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Limitations

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The temperature of the panels was not considered in these calculations. Environmental issues, such as dust gathering on the PV panels were also not accounted for.

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