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How many solar panels do i need for 1,000 kwh per month

For this exercise we will define an estimated value of energy consumption for a house, a building and a factory, and it will be assumed that all the assembled systems are on-grid systems.

How many solar panels do i need for 1,000 kwh per month

We have known for several years the environmental benefits and carbon footprint reduction of photovoltaic solar panels, but there are still many doubts about their operation and profitability. One of these doubts is knowing how many solar panels I need either for a house, a horizontal property building or even in a factory. That is why today we will learn how to perform simple calculations to define the number of solar panels to install for self-consumption.

How many solar panels do I need for my house?
In the case of a single-family home, it is reasonable to establish an average energy consumption of 1,000 kWh/month (considering that in the United States a home can consume an average of 3,000 kWh/month, in Spain about 900 kWh/month and in other Spanish-speaking countries, about 500 kWh/month).

For a consumption of 1,000 kWh/month, the calculation that can be approximated is that of a ratio of 1 kWp installed for every 8 m2. This relationship emerges from the following analysis:

Currently a commercial panel has a capacity of 400 Wp. Each panel can occupy 2 m2. In this sense, to generate 1 kWp (1000Wp), 2 and a half panels are required. Which implies an estimated area of ​​5 m2.

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This increase in the area must always be considered in any installation for the corridor of the photovoltaic solar system maintenance personnel, which is equivalent to the safety distance between the edge of the building (chimneys, antennas, or other installations that can be found on the roof ) and the panel closest to it.

For this reason, another 3 m2 are added for each installed kWp, so that each installed kWp requires 8 m2 of space for its installation.

Now, although the dimensioning of a photovoltaic solar system in terms of panels required for mounting in a house depends on local regulation, most countries have developed regulations that imply that, even when you are not consuming the energy generated by the system, the excess energy injected into the network (that is, the energy generated by the system minus the energy you are consuming at the moment and which exceeds consumption) are remunerated.

For example, in Spain it can be between 5 and 7 cents €/MWh. However, if you generate more energy than you consume per month, the remuneration ceases to be at prices such as the one mentioned and the remunerated value is unfavourable. For this reason, the solar system should be designed to cover the total electrical energy consumption per month at the time the project is dimensioned, considering possible consumption in the near future.

In the case of a home that requires 1,000 kWh/month, first the maximum capacity for consumption that can be installed on the roof must be identified, and then it must be validated if the available roof area is sufficient to install the calculated capacity for consumption.

For example, to calculate how many solar panels we need, we use the following data and formulas:

  1. Monthly consumption: 1,000 kWh/month
  2. Capacity by consumption: 1,000 kWh/month / 30.41 days/month / 3.8 HSP/day = 8.2 kWp
  3. Area required for installation would be: 8.2 kWp * 8 m2/kW = 65.6 m2

Thus, the number of panels required will be:

8,200 Wp / 400 Wp/panel = 20.5 panels, but to ensure total supply it is close to the top unit, which would be 21 panels.

You can apply this calculation even to know how many solar panels you need for a country house, a motorhome or to implement air conditioning or a solar aerothermal system.

The previous result implies adjusting the capacity to 8.4 kWp and a required area of ​​67.2 m2 when going from a requirement of 20.5 panels to one of 21 panels (since 21 panels x 400 Wp/panel = 8,400 Wp).

After identifying this number, it must be validated whether the house actually has the aforementioned roof or floor area. It is recommended that this validation be carried out by a company specialized in this type of assembly, since in addition to validating the available area, safety conditions for its assembly must be evaluated.

If the area is less than the required area, the capacity of the system will be limited to the area available for mounting. If it is higher than required, then the installation can be carried out according to the results obtained.

You should consider:

  • The average number of days in a month: 365 (days/year) / 12 (months/year) = 30.41 days/month.
    The HSP (Peak Solar Hours) of the region where the photovoltaic solar system will be installed. This number depends on the region where the project is running. Peak Solar Hour (PSH) could be defined as a unit responsible for measuring solar irradiation and defined as the time (in hours) of a hypothetical constant solar irradiance of 1,000 W/m². The exercises in this article assume a value of 3.8 HSP.

How many solar panels do I need for a building?
In the case of a building, which is usually located in the city and where the installation area for solar panels is reduced to the area available on the roof, it is not feasible to install a photovoltaic solar system that allows power to be supplied to each home or each office in the.

For this reason, the installations are carried out solely for the purpose of supplying energy to the common areas of the building and economically benefiting the horizontal property or the administration of the building. For this case, an average consumption of 13,000 kWh/month will be assumed.

Carrying out the procedure explained above, the calculations would be:

  1. Monthly consumption: 13,000 kWh/month
  2. Consumption capacity: 13,000 kWh/month / 30.41 days/month / 3.8 HSP/day = 112.5 kWp
  3. Area required for the installation of this capacity: 112.5 kWp x 8 m2/kW = 900 m2

With these data, the number of panels required will be:

112,500 Wp / 400 Wp/panel = 281.2 panels; to ensure total supply, it approaches the upper unit, which would be about 282 panels

The previous result implies adjusting the capacity to 112.8 kWp, and a required area of ​​902.2 m2.

The available area on the roof of a building can usually be between 600 m2 and 2000 m2. If the area is 600 m2, we will have a limitation for the assembly of the system, and the capacity to be installed will go from 112.5 kWp to 75 kWp. This value results from the following operation:

Capacity per available area: 600 m2 / 8 m2/kWp = 75 kWp

It is worth mentioning that usually the roof of a building is made of concrete, which is why it is unusual for there to be complications with the roof’s ability to support the weight of the panels and the people who intend to install and maintain them.

How many photovoltaic solar panels does a factory need?
In the case of a factory, the case of a company that works from Monday to Saturday will be assumed, and whose energy consumption will be assumed for this year at 1,000,000 kWh/month.

Remember that the capacity of the resulting photovoltaic solar system for a consumption like the one mentioned in this example may be large enough that it does not have the same energy surplus compensation benefits as those mentioned for a single-family home or a building.

For this reason, the calculation is made only to guarantee the supply of energy during daylight hours, and that usually, for ease of calculation, is close to 30% of the total consumption.

It is also important to mention that in a project with the characteristic of non-consumption on Sundays, all the energy generated by the system on those days will be surpluses that will be remunerated, but with a value that is surely not representative compared to the value of the kWh that is replaced in daylight hours and paid to the power supply company. That is why it is convenient to use the 30% factor.

Carrying out the procedure explained above, the calculations would be:

  1. Monthly consumption: 1,000,000 kWh/month
  2. Capacity by consumption: 1,000,000 kWh/month * 30% / 30.41 days/month / 3.8 HSP/day = 2,596 kWp
  3. Area required for the installation of this capacity: 2,596 kWp x 8 m2/kWp = 20,768 m2

In this case and with the information we have, the number of panels required will be:

2,596,000 Wp / 400 Wp/panel = 6,490 panels

In this case, the roof area and the capacity of the structure to support the weight of the panels and that of the people who will eventually maintain the system become critical variables that will limit or enable the feasibility of the project, so always it is worth requesting a specialized company to do a project conceptual engineering.

It is also recommended that the project not only have the point of view of a single company but that at least 3 companies be called to compare their points of view, and make a decision based on the knowledge that the suppliers can contribute to the team that go to decide the acquisition.

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