Sunny power station for a house 200 m2 with your own hands
Often on the network, reports of the struggle for ecology, the development of alternative energy sources. Sometimes they even conduct reports on how a solar power plant was made in an abandoned village so that local residents can use the benefits of civilization not 2-3 hours a day while the generator works, but constantly. But all this is somehow far from our lives, so I decided to show and tell how it works and how the solar power plant for a private house works by my example. I’ll tell you about all the stages: from the idea to the inclusion of all devices, as well as share the experience of operating. The article will turn out rather big, so who does not like many letters can see the video. There I tried to tell the same thing, but it will be seen how I myself collect all this myself.
Starting data: a private house with an area of about 200 m2 connected to electric networks. Three.phase entry, with a total capacity of 15 kW. The house has a standard set of electrical appliances: refrigerator, TVs, computers, washing and dishwashers and so on. The power grid does not differ in stability: the record I recorded. a shutdown of 6 days in a row for a period of 2 to 8 hours.
What I want to get: forget about the interruptions in electricity and use electricity, no matter what.
What bonuses can be: to maximize the use of the energy of the Sun so that the house priority eats solar energy, and the deficiency is made from the network. As a bonus, after the adoption of the law on the sale of electricity to the network, to start compensating for part of their costs, selling excess production in a common power supply.
Methods to calculate the power of solar batteries for the home and giving only two. It is recommended before installing solar panels, write down data on spent energy for several months in order to have an average value.
Or, to calculate the total power of household appliances that you constantly use. It is in technical documents for electrical appliances. You can find it on the Internet, setting the name of the model in the search bar.
Knowing the power of the devices used in the house, it should be multiplied by the time during which they work within a day. All data obtained are added up. This will be the figure for orientation.
If it is planned to install an inverter with a controller, they are also taken into account when calculating the final power of solar panels installed in the house or cottage.
Online male battery calculator
Insolation data were provided by the NASA server, the history of measurements has been conducted since 1984 and is the most reliable information in the world today.
one. The selection of the region first you need to choose on the map the region of placement of a solar power plant.
The calculation of the load when choosing solar batteries is extremely important to correctly calculate the power of power consumption. To do this, in the calculator you are invited to indicate the electrical devices that you will use. Indicate their number, power, as well as the time of work during the day. If the list we have proposed does not need the device you need, you can use the “Other Device” item.
The choice of equipment and options from the drop.down list first select the solar modules that you plan to use or already use, do not forget to specify their number. If the proposed list does not have the necessary solar module, select “I have a different solar module”, then to fully determine the cost of a solar power plant, you have to select a network inverter from an drop.down list and, if necessary, additional options such as: fasteners (a system of fasteners of solar modules on plane), consumables that may be required when installing the sisetm (metizes, thermal shrinkage, screed, etc.).
Getting data, our calculator will automatically calculate and show the optimal angle of inclination (“optimum”) for maximum average production per year, as well as the optimal winter and summer angle, which will be useful in the case of using the rotary mechanism or operation of the power plant at a certain time of the year (for example Only in the summer, in this case, you should focus on the angle “Summer”). If for some reason you do not want to use the optimal angles proposed by the system (for example, you plan to mount the modules on the roof of your house, and the angle is determined by the existing design), it is possible to set an arbitrary (the angle you need). When the angle is changed, the development data will be recalculated automatically.
How many solar batteries do you need, how to calculate solar batteries for a private house
There is no unequivocal answer to this question. It all depends on the amount of energy consumed in your house: refrigerator, kettle, microwave, TVs, washing machine, computers, air conditioning, dishwasher and so on.
Based on the data on energy consumption, inclination and roof area, choosing the power of solar panels, the necessary number of solar batteries for SES is calculated. How many solar batteries are needed for a home or apartment. a specialist will definitely be able to answer, only after leaving for an object.
How to calculate how many solar batteries are needed
To calculate the required amount of light batteries for the house, you need to build on several factors:
- Roof size;
- The amount of electricity consumed per month;
- The amounts that are ready to invest in the project;
- The power that is spelled out in the contract for the use of electric energy (the power of the station should not be more than in the green tariff agreement).
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If you need a station exclusively to replace your own consumption, it is necessary to build on the average monthly energy consumption by the house. Before buying, the calculation of power must be made, especially if the task is to recoup the solar station as quickly as possible, you need to install the maximum possible power of the station.
How to independently calculate solar panels and their development
We will use a simple formula with which you can approximately calculate how many solar panels for the house you need. To do this, you need:
- Know the area of the roof, for example, take 50 m²;
- The area of one solar battery, we take 1.63 m²;
- And the power of the solar panel. take an average market for 275 watts.
Now how to calculate the solar batteries for the home using the formula:
- We divide the area of the roof into an area of one panel: 50/1.63 = 30.67, round the smaller direction, we get 30 solar panels.
- Next, we multiply the resulting number of panels by the power of one panel: 30275 = 8250 W, twist the smaller way and leaves 8,000 watts or 8 kW per hour.
We get a solar power plant with a capacity of 8 kW with a maximum number of solar batteries of 30 pieces, which will cover a roof with an area of 50 m².
Once again we remind you that the above numbers are just an example of calculating the approximate number of solar batteries for a private house. The formula was not taken into account: the type and angle of inclination of the roof, loss of efficiency in the inverter and battery, seasonality, etc.P.
What electrical appliances are taken into account when calculating SES indicators for a private house
Detailed attention to consumer power should be given in the case of designing autonomous systems and backup systems. In such systems, the generating equipment should cope with the load from consumers, even have a certain power supply for the duration of the starting processes.
Insufficient equipment capacity in such systems can lead to an emergency power outage during overload, and, even, to its failure.
It is worth considering all powerful equipment in the house: electric boilers, electric warm floors, deep and circulation pumps, compressors of refrigerators, etc.
In the case of a network solar power plant, it is more important to take into account the consumption of electrical appliances, since all of the consumed capacity is taken away from the volume of electricity of the manufactured power plant, and the lack of power inverters is covered with consumption from the network.
How to improve electricity generation indicators
According to the California study, regular batteries for solar power plants allows you to produce more by 12 % of electricity. To develop the maximum number of solar elements, the SES is installed at a certain angle, taking into account the structure of your roof. The optimal angle of inclination is 45 °.
We will be glad if this material made it possible to find out how to calculate solar panels on your own. But we would recommend contacting specialists to conduct such calculations.
Where can I buy
You can purchase solar panels both in a specialized store and online in the online store. In the second case, the budget option for purchasing products on the Aliexpress website deserves special attention. For some panels, there is an option to ship from warehouse B, they can be obtained as quickly as possible, for this when ordering, select “Delivery from the Russian Federation”:
Heliobatary design is not the only factor that determines the operational indicators of the complex. External factors interfere in the process that reduce the capabilities of the complex. They affect the operation of the equipment one by one and together, reducing the efficiency and reducing the indicators of the heliocation.
The power of the solar battery is the amount of electricity, which it is able to issue per unit time. This is the final value, that is, calculated by the maximum value and having a certain limit. It is known that the solar permanent is 1 kW per 1 m². This value is measured in certain conditions, indicates the amount of energy falling on the earth’s surface on a sunny day at a temperature of 25 ° and constantly vertical drop to the surface. In practice, obtaining a full calculated volume of energy is impossible.
The efficiency of solar panels is limited and does not exceed 24 %, therefore, the maximum power obtained from 1 m² of the receiving surface may be 0.24 kW. This is in ideal conditions and with constant correction of the surface position relative to the sun. In practice, there are no such conditions. Weather, climatic and seasonal conditions interfere in the situation. Entire cloudy weeks are possible, the duration of daylight hours in the summer and winter period is significantly different.
In addition, the temperature also affects the ability of solar elements to produce energy. its production drops significantly as soon as the temperature rises above 25 °. This means that on a clear summer day, when the power of solar batteries per square meter should be maximum, it will not be possible to get the expected result due to severe heating of photocells. Therefore, when calculating a solar power plant, it is necessary to make amendments to seasonal conditions, duration of the day and other natural factors.
The next factor that must be taken into account when performing the calculation is the degradation of heliopanels. This indicator in different models differs, there are samples that save up to 90 % of working qualities even after 20–25 years of operation. However, in most panels, degradation occurs evenly and proportional to the duration of use.
In addition, the calculation of the number of solar panels must be done taking into account losses on additional equipment. the inverter has an efficiency of about 92–96 (and this is one of the best models). In addition, the losses on the battery and controller, which reach 40 % and also reduce the total parameters of the complex are inevitable. The devices themselves consume energy to power their own boards. Therefore, a complete and accurate calculation of solar panels is an extremely complex task, requiring experimental confirmation.
Sunny station for the house 5.525 kW. option 6
Alteko offers to consider the typical solution “Autonomous Solar Power Plant for the House 5.525 kW” for autonomous or reserve power supply of an object based on the renewable energy of the Sun.
The proposed equipment is the base and can be changed for specific requirements to ensure specified parameters.
Characteristics of the solar station:
Power of solar panels. 5525 watts; The power of the connected load (inverter power) is 5.0 kW; Type of solar panels. monocrystalline; El.
energy for a month (in the winter). up to 238 kWh; El.energy for a month (in the summer). up to 816 kWh; El.energy for 8 months (from March to October). up to 5376 kWh; El.
energy for 12 months. up to 6248 kWh; Battery capacity. 420 Ah, 48V (20.2 kWh).
The composition and cost of the solar station (option 6)
The cost of the solar station takes into account: the station connection scheme and free consultation of those.specialist, in the case of independent (remote) installation.
Structural scheme of the solar station
It usually looks like this: installation of panels on the roof
300 watts Solar system for home calculations │ Battery Inverter Ups requirement
- Advantages of private solar power plants:
- Savings at the stage of purchase of land for construction;
- Savings at the stage of connection to electric networks;
- Complete autonomy and lack of payment for electricity;
- The ability to sell electricity to the network at the “green tariff”.
The initial data for calculations
Now let’s look at how to calculate solar batteries? The main figure necessary for calculations is the general energy consumption for a certain period. If the panels are installed in an electrified country house, then electricity consumption can be determined by the counter. However, if the power supply is connected for the first time, it is necessary to make a list of all available consumers indicating the power of each of them.
For example, the refrigerator consumes 350 W/h. It will consume about 1 kW/h per day, and within a month. about 30 kW/h. In the same way, you need to calculate the electricity consumption in lighting and other devices.
The received numbers are added up and first the general daily energy consumption is determined. Further, the result is multiplied by the number of days in a month, which will give preliminary value. For example, electricity consumption is 100 kW/h. This figure will be relative, since it should be added another 40% to losses in the battery and when working inverter.
Thus, the total electricity consumption per month will be 140 kW/h. A day is 140: 30: 7 = 0.67 kW/h. Therefore, panels with a minimum power of 0.7 kW are needed. However, they will be enough only in good weather in the summer and partially in the spring and autumn. It is necessary to take into account cloudy days, which are often observed in the summer months. In this regard, it is required to increase the number of panels by at least doubles, otherwise the electricity will be intermittent.
The maximum effect of the solar system is obtained only if the work of all the components of the parts and components is consistent. First of all, it is necessary to correctly calculate the batteries based on the source data, because it is precisely these calculations that will depend on the efficiency of the entire energy installation.
- Panels. When ordering from the manufacturer (Solarworld, Germany), one panel costs 350. 120 panels 350 = 42.000
- Fasteners. Usually, when mounting on a tin roof, guide rails and a designer are used. an aluminum alloy, bolts. stainless steel. In terms of one panel, 3 meters of rails, 10 bolts with laying, 4 bolts with a half.beam are consumed. Fasten costs. 6.000
Theoretically, in ideal conditions, one panel should issue approximately 220-230W per hour (in terms of 220 volts familiar to us). Below are the graphs that the control unit in the inverter leads, you can monitor them remotely.
In the last schedule, it should be noted that the system turned off for two days for a while, and the first three days of the month and the last two are absent.
In a stable sunny summer month, with a long daylight day, such a farm will give out a maximum of 4500-4700kw. Knowing these numbers, the profitability of the system can be calculated, given the tariffs for electricity.
At the same time, it must be taken into account that the farm is assembled without batteries, their presence would increase the total cost of the system, the payback time, respectively, also.
Thus, I can’t get to payback in 2-3 years. 10 years-more or less real term.
How to calculate your own assessment of the solar panel
For those who are interested in how we appreciated these figures for energy consumption and the required number of solar batteries, here is a breakdown. If you want to understand how much energy you need, start by how many kilowatt hours (KWh) you use in the year. Most utilities provide you with the total electricity consumption over the past twelve months in a monthly extract or receipt for payment. To offer some perspective, one kW-hour. it is 1000 watts of energy used per hour. Thus, if your house has 20 bulbs, and they all use 50 watts each, then when each light bulb in your house will be consumed 1 kWh electricity within an hour. According to the latest data, in 2016 the average American family used 897 kWh per month. In other words, the middle family consumes a little less than 11,000 kWh per year.
To find a range for the number of solar panels, we compared the coefficients of the production of solar panels, the highest and lowest. Then we took 11,000 kWh and divided them into the corresponding coefficients, and then divided this number. on 250 (typical panel power). This calculation gave us the maximum and minimum for the average number of panels that the homeowner will need.
How many kWh your solar batteries can produce?
The amount of energy (kWh) that your solar energy system can produce depends on how many sunlight falls on your roof. The amount of sunlight that you receive in a year depends on where you are in the country and on the time of year. There are more sunny days in the south, respectively, than on the western or even northern part of the country. But anywhere you can produce enough energy to satisfy your energy needs! If you live in a area where there is less sunlight, you just need to install a system that will contain more solar panels.
Two households of comparative size in the Krasnodar Territory and for example in the Novosibirsk region consume the average volume of electricity, about 10,400 kWh per year. The Krasnodar household needs a system of 7.0 kW to cover 100% of its energy needs. For comparison, a comparable family in Novosibirsk needs a system of 8.8 kW to meet their needs for energy. Solar batteries in the Krasnodar Territory will need less than in the system in Novosibirsk, but are able to produce the same amount of energy, because they are subjected to more sunlight annually. Homeowners in less solar areas can make up for this discrepancy, simply using more effective panels or increasing the size of their solar power plant, as a result of which there will be slightly more solar panels on their roof!