DIMENSIONING OF A PHOTOVOLTAIC SYSTEM

There are many pages and articles on the internet where they teach you how to calculate how many solar panels you need for your home, today we are going to show you how to dimension an autonomous photovoltaic solar installation step by step

DIMENSIONING OF A PHOTOVOLTAIC SYSTEM

In the case of an  installation of autonomous photovoltaic solar energy  (isolated from the network), it is essential to have a correct sizing both to be able to supply the energy demand with guarantees   , as well as to limit the  economic cost  of the installation.

As an example, we will take the need to electrify a  house  without an electrical connection to the network  in a rural area, which will be used by a family of 4 people on weekends.

Let’s see the calculations step by step:

First step: Calculation of estimated consumption

We establish for the example case the necessary basic equipment that will consume energy:

Bulbs: 4 units x 4 hours x 60 Watts (100%) = 960 Wh

Television: 1 unit x 3 h x 70 W (100%) = 210 Wh

Laptop: 2.5 hx 60 W (100%) = 150 Wh

Refrigerator: 24 hx 200 W (50%) = 2400 Wh

Microwave: 0.5 hx 800 W (100%) = 400 Wh

In this section, you will have to estimate the consumption for your specific case. The necessary consumption could be estimated here for other types of installations, such as the demand for self –  consumption  to partially cover the needs of an  installation connected to the network  or an installation designed to supply a  recharging point  for a bike, motorcycle or electric car, to charge the batteries, etc.

Later we will make more specific articles for this other type of cases, today we are going to focus on our example for an  isolated house .

So, if we add the different partial consumptions, we obtain the total estimated consumption for our example house:

Total consumption per estimated day (Cde) = 4120 Wh / day

We apply a   75% installation performance to calculate the  total energy needed to supply the demand:

Total energy required (Ten) = Cde / 0.75 = 5493 Wh / day

Second step: Solar radiation available

To obtain the  incident solar radiation , tables with already existing estimates can be used. A good source of these estimates is the PVGIS application (Photovoltaic Geographical Information System – European Commission, Joint Research Center), which has an online platform from which insolation data for all of Europe can be obtained easily and quickly.

As an example, you can do it with any city

Assuming that our installation is in Granada, using the PVGIS application   we obtain the following values:

Latitude: 37 ° 10’38 “North
Longitude: 3 ° 35’54” West
Nominal power of the photovoltaic solar installation: 1kWp
Inclination of the modules: 35deg.
Orientation of the modules: 0deg.

Month	Ed	Em	Hd	Hm
January	3.45	107	4.35	135
February	4.11	115	5.25	147
March	4.70	146	6.21	193
April	4.53	136	6.05	181
May	4.76	148	6.49	201
June	5.11	153	7.14	214
July	5.26	163	7.49	232
August	5.18	160	7.34	228
September	4.69	141	6.46	194
October	4.39	136	5.88	182
November	3.63	109	4.66	140
December	3.38	105	4.27	133
Total	4.43	135	5.97	182

Where:

Ed: Average daily energy production of the system (kWh)
Em: Average monthly electricity production of the system (kWh)
Hd: average daily sum of global irradiation per square meter received by the system modules (kWh / m2)
Hm: average sum of global irradiation per square meter received by system modules (kWh / m2)

The most unfavorable month of radiation, we observe that it is in December with 4.27 kWh · m2 / day. So we will  size the installation for the most unfavorable monthly conditions of insolation , and thus we make sure that we will cover the demand throughout the year.

Once we know the incident solar radiation, we divide it among the incident solar radiation that we use to calibrate the modules. (1 kW / m2), and we will obtain the amount of  peak sun hours (HSP) . For practical purposes in our case this value does not change, but we will use the concept of HSP (peak sun hours) which is the equivalent number of hours that the sun would have to shine at an intensity of 1000 W / m2 to obtain the  total insolation  of a day, since in reality the sun varies intensity throughout the day.

HSP = solar radiation tables / 1kW / m2 = 4.27 HSP

Third step: Calculation of solar panels or panels needed

We will perform the calculations to establish the number of modules (plates or solar panels) according to the most unfavorable radiation conditions. To perform this calculation we have chosen modules of 180 W. This data is given in the  technical characteristics of the modules  chosen according to each model and manufacturer.

1. For installations of daily  use we will use the formula:

Number of modules = (energy needed) / (HSP * work efficiency * module peak power)

The  working efficiency  takes into account losses caused by possible fouling and / or deterioration of the photovoltaic panels (normally 0.7 – 0.8).

Number of modules for installation of daily use:

Nmd = (5493) / (4.27 * 0.8 * 180) = 8.9 Rounding 9 modules

2. For weekend facilities we will  use the formula:

Number of modules = (3 * energy needed) / (HSP * work efficiency * 7 * module peak power)

Number of modules for installation for weekend use:

Nmfd = (3 * 5493) / (4.27 * 0.8 * 7 * 180) = 3.8 Redoing 4 modules

As our example case is for a house that is used on weekends, we will need four modules of 180 W each. Taking into account that the consumption needs that we have established are very basic, if we introduce greater consumption in the first section we will find a larger number of plates.

With the chosen modules of 180 Watts peak (Wp), we will obtain a total solar installation of 720 Wp (4 x 180 Wp).

Taking into account that the  modules work at 12V , if we want an installation that works at  24V , we can make an association in series of groups of two plates and then these two groups of two plates in series, associate them in parallel. The  operating voltage  will depend on the system of accumulators that we choose.

Fourth step: Accumulator capacity

To design the capacity of  accumulation batteries , we will first have to establish the  desired autonomy   in case of having unfavorable days without insolation due to abundant cloudiness.

In the case that concerns us, for weekends the maximum necessary autonomy can be established in 3 days (Friday, Saturday and Sunday). In  electrification of rural houses  for daily supply could be set between 4-6 days, taking into account that this value can be reduced in the case that we have a  generator set of reinforcement .

Battery capacity = (energy required * days of autonomy) / (Voltage * depth of battery discharge)

The  depth of discharge  depends on the type of battery chosen. These values ​​range from 0.5 to 0.8. You can consult these values ​​in the technical characteristics for each model and manufacturer. In our case, we will choose a battery that can tolerate a discharge of up to 60% (0.6).

Accumulation capacity = (5493 * 3) / (24 * 0.6) = 1144.38 Ah (c100)

The value c100 indicates that the capacity of the battery will be that provided by 100 h charge cycles, which is the load frequency normally established in rural electrification.

The selection of  the accumulation system  requires different checks so that the system lasts and has an optimal performance. The accumulation systems need a minimum load intensity to ensure that the batteries charge correctly and prevent them from having a shorter life than expected.

This article is intended to be a basic example of the calculation of the parameters necessary to perform an installation, but once we know the necessary capacity for our installation, we recommend contacting specialists to know more details or information on the technical characteristics of a specific system or manufacturer of accumulators.

You can access our directory of renewable energy companies and professionals to find installers, manufacturers or distributors of photovoltaic solar systems and accumulation batteries near your locality, and make consultations without obligation.

Step five: Selection of regulator and converter

Finally, we would only have to choose a  charge regulator  and a  direct current to AC converter  to be able to have 220 V AC in our house suitable for any type of appliance or household appliance.

The charge regulators are determined by the maximum working intensity and by the voltage in which we have designed our installation.

The power of the DC / AC converter will have to be chosen depending on the sum of all the nominal powers of the consuming equipment multiplied by the coefficient of simultaneity of use of these. (usually values ​​ranging from 0.5-0.7). In our case, the total estimated power is 1360 W

Converter power = 1360 * 0.7 = 952 W

So, with a 1000 W converter it would be enough for our example, as long as we really use only the devices initially considered. We can always establish a higher power in case some other household appliance of higher consumption is used.

Autonomous or isolated installations of the grid and other uses of solar photovoltaic energy

In this article, we have shown you an example to size a  solar photovoltaic installation isolated from the network . These facilities are dimensioned reducing to the minimum the consumptions to realize to have an installation of affordable cost, but this supposes to do without some elements of greater consumption and power. For this reason, in this article we have not taken into account washing machines or ovens, etc. since it is the assumption of a house with a very basic electrical installation for second home use.

Although the first step to save with renewable energy is to try to reduce consumption, it is not always possible to perform such a basic installation as the one proposed in the example. For other types of homes or buildings, there are other options to save with photovoltaic solar installations. One option is to make a  solar installation for self-consumption connected to the electricity grid . In these cases, the installation is complementary and allows you to save but at the same time remain connected to the network.

You can see in our article “Zero injection an alternative for self-consumption” how these systems work. You can also see where modules are necessary in our article “Solar photovoltaic kits for electrical self-consumption What are they? How do they work? “Or you can calculate the number of plates needed according to your consumption in our photovoltaic solar energy calculator.

Another option for which autonomous systems such as the one explained in this example can be very useful is to supply, for example, a recharging point for batteries for bicycles, motorcycles or electric cars. We will discuss these cases in future articles. We hope that the article is of your interest. You can leave your comments and we will try to solve your doubts or queries!