All the calculation is done on the 3rd day of September thus the day number is 246. Latitude angle is 51.43˚. Declination angle can be calculated from equation 2 as following:
To find out the sunset angle equation 5 can be used as the following:
The daily extraterrestrial radiation on a horizontal surface has been calculated using the following:
To find the clearness index, equation 8 can be applied. However, before, the value of the average radiation on a horizontal surface, should be calculated in J/m2, thus, the value in Wh can be multiplied by 3600 to be 10.08 MJ/m2. Now, the clearness index can be found as following:
Now, to find the percentage of diffused radiation to the global radiation, equation 10 or equation 11 can be applied depending on ws. In the studied hour case equation 10 should be used.
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The diffused radiations can be found by multiplying this ratio by the global irradiation on a horizontal surface.
Sunrise hour and sunset hour has been calculated for this day to be 6:16 and 19:43 consequently. The length of this day is found to be 13 hours and 27 minutes.
The solar noon takes place at 13:00. (Note that all those values are at the local time, and consider the one hour shift forward to start at 25th of March and ending at 28th of October).
To calculate the hour angle, as mentioned before, the hour angle is -15˚ for each hour before noon and +15˚ for each hour after the solar noon. In the case of the example, the hour intended to find the power output at, is between 12:00 and 13:00, local time. While the time used is local time and is shifted one hour forward, the solar noon should be calculated according to one hour earlier. The solar noon occurs at 12:00 exactly, 13:00 at the local time, thus the hour angle between 12 and 13 becomes the hour angle between 11 and 12. It is thus calculated to be -15/2, which is -7.5˚.
Now, Collares-Pereira's equation can be used to break the daily radiation into an hourly global irradiance. The equation applied in the hand calculation is as follows:
So rt is found as follows
Constant (a) and constant (b) is found as follow, noting that the angles are used in radians. For ws = 98.81˚, the angle in radians equals to 1.724, thus:
Omar Hamdan | Kingston University London And (b) found as follow:
So, rt can be found to be:
The hourly diffused irradiance can be broken down using the Jordan and Liu equation:
So,
And:
The final irradiance on a tilted surface can now be found using equation 19, introduced by Jordan and Liu (Duffie and Beckman, 2006) mentioned. Considering the slope angle of the PV panel to be 51˚, this can be used as following:
Where thus,
Omar Hamdan | Kingston University London And,
Then, this value can be found in Watt as the following:
This is the maximum power that can be extracted from the sun at the latitude mentioned and the specifications stated. Now the electrical power output depends on the efficiency of the PV panel and its efficiency variation with the ambient temperature. Furthermore, the electrical connection and the electrical losses will take an effect on the power output mentioned. Those calculations were done using a simulation program which will help to design the system's specification more accurately.
Finally, it is important to show the effect of the temperature on the PV panel's efficiency, thus calculations were carried out, whilst consider the power output. It can be presented for the case study stating that the efficiency is going to change according to the internal temperature Tc of the PV cells. Each material and each manufacturer type PV will vary differently according to its internal temperature. The ruling coefficient for this change is the , the temperature coefficient for efficiency.
The PV used in this project has a maximum power point voltage equal to 24 Volts.
The temperature coefficient for open circuit voltage is equal to -104 mV for each ˚C degree. From this data the efficiency temperature coefficient can be found, due to the relationship between the voltage and efficiency being linear. Thus the efficiency coefficient is:
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The relationship between the efficiency and the internal temperature is stated in equation 26. Applying it to the studied hour will need to find the value of Tc which in its place will vary according to the ambient temperature. The average ambient temperature of September is 16.7 ˚C. The reference PV efficiency is found under Standard Testing Conditions (STC), while under those conditions, the temperature is known as Nominal Operation Cell Temperature (NOCT). The slope angle that used in calculation does not equal to the latitude angle minus the declination angle. In this case a correction factor should be used by applying equation 28, the optimal angle for Cf is equal to 44.0421˚, so:
The PV selected has a NOCT equals to 47.5 ˚C. Equation 26 can be applied to find Tc as following but the first part, excluding Ta, has to be multiplied by Cf as following:
Now, efficiency equation that relates the temperature with the reference temperature can be applied:
Finally, the final power (W/m2) can be found. The power produced between 12:00 and 13:00 on the 3rd of September, when using the specified PV panel, is:
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The system will produce, at the specified time, 22708.4047 Wh when the optimal area is used. The optimal area is calculated for the site to be 408 m2. (see the area assessment section for more details about the system size and area assessment).