Photovoltaic Solar panel installation
As I have seen that the examples of calculation cases help to solve many doubts about how to implement photovoltaic installations, today I am going to show you how to design an isolated autonomous photovoltaic Solar panel installation.
To dimension an autonomous photovoltaic installation (isolated from the grid), it is important to establish a balance between the energy demand and the economic cost of the installation.
Calculation of energy consumption
This data will be obtained by adding the electricity consumed by each of the electrical appliances of the house and multiplying it by the hours that are used per day.
||320 wh / day
||480 wh / day
||175 wh / day
||120 wh / day
||00 wh / day
||22400 wh / day
||3695 wh / day
Total consumption per estimated day (Cde) = 3695 Wh / day
We apply the performance of the installation of 75% to calculate the energy needed to supply the demand:
Total energy required (Ten) = Cde / 0.75 = 4926 Wh / day
Solar radiation available
To obtain the incident solar radiation, tables with already existing estimates can be consulted. A good application to obtain this data is PVGIS (Photovoltaic Geographical Information System – European Commission, Joint Research Center) that has an online platform from which isolation data can be obtained for all Europe easily and quickly.
- Other losses (cables, inverter, etc.): 14.0%
- Combined losses of the PV system: 24.5%
- Ed: Average daily electricity production by the given system (kWh)
- Em: Average monthly electricity production by the given system (kWh)
- Hd: Daily average of the global irradiation received per square meter by the modules of the given system (kWh / m2)
- Hm: an Average sum of the global irradiation per square meter received by the modules of the given system (kWh / m2)
- To ensure that we will cover the demand throughout the year we choose the most unfavorable month of radiation, which in our case we see that it is December with 3.87 Kwh – m2 / day.
- HSP = solar radiation / 1KW / m2 = 3.87 HSP
Now we must calculate the number of panels (modules or solar panels) based on the most unfavorable radiation conditions. To make this calculation I have chosen modules of 250 W, with a maximum current of 8.69 A, at 24V. This data is obtained from the technical characteristics of the solar panels chosen according to each model and manufacturer.
Here I present two different scenarios to calculate the number of solar panels that our photovoltaic installation needs to supply the established energy demand
For installations of daily use we will use the formula:
Number of panels = required energy / HSP * working efficiency * peak power of the solar panel
The work efficiency refers to the losses produced by the deterioration of the solar panels and/or fouling of the same (normally it is 0.7 – 0.8)
The number of panels for daily use installation (NPD):
NPD = 4926 / 3.87 * 0.8 * 250 = 6.36 (7 solar panels)
For weekend facilities we will use the formula
Number of panels = 3 * energy needed / HSP * work efficiency * 7 * peak power of the solar panel
Number of panels for installation for weekend use:
Npfd = 3 * 4926 / 3.87 * 0.8 * 7 * 250 = 2.72 (3 solar panels)
Observe that the energy demand that I have established as an example is to meet some very basic needs if you enter higher consumption in the calculation of energy consumption will result in a greater number of panels.
With the chosen panels of 250 watts peak (Wp), we will have a solar installation of 750 Wp (3 x 250).
The operating voltage will depend on the batteries we choose for our accumulation system.
For our example, we will use a 24 V system, for which we will only have to arrange the three solar panels in series.
Battery capacity = energy needed * days of autonomy/voltage * depth of battery discharge
The depth of discharge depends on the battery that we choose for our accumulation system. Today in the market we find batteries whose depth of discharge ranges between 30% – 80%.
Accumulation capacity = 4926 * 3/24 * 0.3 * = 2052.50 Ah (c100)
Our accumulation capacity will be 2052.50 Ampere-hours (Ah). The value c100 indicates that the capacity of the battery will be supplied by discharge cycles of 100 h, which is the frequency of discharge normally established in rural electrification.
The power of the inverter will have to be chosen based on the sum of all the nominal powers of the equipment multiplied by the coefficient of simultaneity of use of these (these values go from 0.5 – 0.7). In our example, the estimated power is 1320 W.
Inverter power = 1320 * 0.70 = 924 W