How to calculate the solar panels for the house? How many solar panels and batteries are needed for your home.
With the help of autonomous solar installation you can provide energy to all electrical appliances in your home. The main thing is to understand and correctly assess the needs of your household and the capacities that you need to establish.
Components of the home solar system.
Home photovoltaic system, as a rule, consists of 6 basic elements:
We calculate the number of solar cells and batteries in 6 steps
1. Calculation of energy consumption. The first step is the specification, that is, the technical description of the system. First you need to make a list of all electrical appliances in the house, find out their needs and put them on the list.Below are the approximate data on the average power values of some devices. These are rough estimates. In order to calculate the power consumption of the system with the inverter (for AC devices), corrections must be made for each device. The losses in the inverter can be up to 20%. The refrigerator, the compressor at the time of starting consume power 5-6 times more than the passport one, so the inverter must withstand short-term overloads 2-3 times higher than the nominal capacity. If there are many appliances with a high power, then for a cheaper and optimal choice Inverter, it is necessary to provide separate inclusion of such devices during operation.
Table 1. Average power with electrical appliances
2. Determine the amount of solar energy that can be obtained in a given area. Two factors are important here:
- average annual solar radiation,
- average monthly values for the worst weather conditions.
Based on this table, it is possible to select the power of solar panels taking into account actual consumption, except in cases of extremely long periods of bad weather. Using modules of different power (50, 100, 250 W), you can gain power for your own system.
3. Battery capacity selection depends on the need for energy and on the number of panels - on the charging current. AGM batteries need 10% charge current. For a 90W panel, the minimum battery capacity is 60Ah, and the optimum one is 100Ah. It will store 1.2kWh at 12V. 
For systems of consumption up to 1.5 kWh per day, it is better to use 12V batteries and panels. Systems that consume more than 3 kWh per day are advisable to be equipped with a solar generator and a battery with a voltage of 48 V.
The most affordable are car batteries, but they are designed to transfer large currents for a short time. These batteries do not withstand the long charge-discharge cycles typical for solar systems.
Special solar cells have low sensitivity for cyclic operation and low self-discharge. Manufacturers produce batteries with different discharge times. The selected battery should have an energy reserve of approximately 4 days.
To ensure that the battery has served the manufacturer's stated time limit, it should be used in conjunction with a quality charge controller. The charge current is monitored, which decreases with a fully charged battery. The supply of energy is interrupted during discharge to a critical level. 
4. Inverter selection. To use household appliances, alternating current (220V, 50Hz) is used, and for this purpose an inverter must be in the solar system with the battery. It is advisable to use inverters with a sinusoidal output - this is qualitative for devices.
5. Life of components. An important factor is the life of individual components. Photopanels provide for a decrease in productivity of up to 80% in the 20 th year, although they can work for 25 years. Frameworks and fastenings too should be chosen for such term: aluminum or a stainless steel. Batteries have an average service life of 4-12 years (depends on the nature of charge / discharge cycles). Inverters mainly serve 10-15 years, and the warranty period is set for 5 years. 
6. Moral obsolescence and disposal. Little is said about this, but this can not be avoided. All technological innovations and. Every year there are more and more productive and cheap
I welcome you on the site e-veterok.ru, today I want to tell you about how much you need solar panels for home or cottage, private house, etc. In this article there will be no formulas and complex calculations, I will try to convey everything in simple words, understandable for any person. The article promises to be not small, but I think you will not waste your time, leave comments under the article.
The most important thing to determine the number of solar cells you need to understand what they are capable of, how much energy can give one solar panel to determine the right amount. And also need to understand that in addition to the panels themselves will need batteries, a charge controller, and a voltage converter (inverter).
Calculation of solar power
To calculate the necessary power of solar cells you need to know how much energy you consume. For example, if your energy consumption is 100kWh per month (the readings can be viewed on the electricity meter), then you need the solar panels to produce such amount of energy.The solar panels themselves produce solar energy only during the daytime. And they give out their passport capacity only if there is a clear sky and the sun's rays fall at a right angle. With the fall of the sun at corners, power and power generation drop noticeably, and the sharper the angle of sunlight fall, the more power is lost. In cloudy weather, the power of solar batteries falls 15-20 times, even with light clouds and haze, the power of solar batteries falls by 2-3 times, and this all must be considered.
At calculation it is better to take working time, at which solar batteries work almost on all capacity, equal 7 hours, it from 9 mornings to 4 o'clock in the evening. Of course, the panels will work in the summer from dawn to dusk, but in the morning and in the evening the output will be quite small, only 20-30% of the total daily output, and 70% of the energy will be produced in the interval from 9 to 16 hours.
Thus, an array of panels with a power of 1 kW (1000 watts) for a summer sunny day will give 7 kWh of electricity for the period from 9 to 16 hours, and 210 kWh per month. Plus 3kW (30%) for the morning and evening, but let it be a reserve so as it may be partly cloudy. And the panels we have installed permanently, and the angle of incidence of the sun's rays varies, from this naturally the panels will not give out their power by 100%. I think it's understandable that if the array of panels is 2kW, then the output will be 420kWh per month. And if there is one socket per 100 watts, then per day it will give only 700 watts of energy, and a month 21 kW.
It's nice to have 210kWh per month with an array of only 1kW, but it's not that simple
At first there is no such thing that all 30 days in a month are sunny, so you need to look at the weather archive for the region and find out how many roughly overcast days are by months. As a result, probably 5-6 days will definitely be overcast, when the solar panels and half of the electricity will not produce. So you can safely cross out 4 days, and it will no longer be 210kW * h, and 186kW * h
Also you need to understand that in the spring and autumn the light day is shorter and the cloud days are much longer, so if you want to use solar energy from March to October, you need to increase the array of solar cells by 30-50%, depending on the region.
But that is not all, there are also serious losses in the batteries, and in the converters (inverter), which also need to be taken into account, about this further.
About the winter I will not speak until this time is quite deplorable in the development of electricity, and then when there is no sun for weeks, no array of solar cells will help, and it will be necessary either to eat from the grid during such periods, or to put a gasoline generator. The installation of a wind generator also helps, in the winter it becomes the main source of electricity generation, but if, of course, there are windy winters in your region and a sufficient wind generator.
Calculation of battery capacity for solar panels
The solar power station inside the house looks like this
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Another example of installed batteries and a universal controller for solar panels
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The lowest battery capacity, which is simply necessary must be such to survive the dark time of the day. For example, if you are consuming 3 kWh of energy in the evening and up to the morning, then there should be such a store of energy in the batteries.
If the battery is 12 volts 200 Ah, then the energy in it can fit 12 * 200 = 2400 watts (2.4kW). But the batteries can not be discharged 100%. Specialized batteries can be discharged to a maximum of 70%, if more then they quickly degrade. If you install conventional car batteries, they can be discharged at a maximum of 50%. For this reason, it is necessary to put batteries twice as long as required, otherwise they will have to be changed every year or even earlier.
Optimum battery capacity this is the daily energy reserve in the batteries. For example, if you have a daily consumption of 10 kW * h, then the working capacity of the battery should be exactly this. Then you can easily survive 1-2 cloudy days without any problems, without interruption. At the same time on ordinary days during the day the batteries will be discharged only 20-30%, and this will prolong their short life.
Another important thing to do this is the efficiency of lead-acid batteries, which is approximately 80%. That is, the battery with a full charge takes 20% more energy than it can then give. The efficiency depends on the charge and discharge current, and the more the charge and discharge currents, the lower the efficiency. For example, if you have a 200Ah battery and you connect an electric kettle to 2kW through an inverter, then the battery voltage will drop dramatically, as the battery discharge current will be about 250Amper, and the efficiency of energy release will fall to 40-50%. Also, if you charge the battery with a large current, the efficiency will drop sharply.
Also, the inverter (energy converter 12/24/48 in 220V) has an efficiency of 70-80%.
Given the loss of energy from solar cells in batteries, and on the conversion constant voltage in the variable 220V, the total loss is about 40%. This means that the storage capacity of batteries should be increased by 40%, and so increase the array of solar cells by 40%to compensate for these losses.
But this is not all the losses. There are two types of battery charge controllers from solar panels, and they can not be avoided without them. PWM (PWM) controllers are simpler and cheaper, they can not transform energy, and therefore solar panels can not give up all their power to the battery, at most 80% of the rated power. But MPPT controllers monitor the point of maximum power and convert the energy by lowering the voltage and increasing the charging current, ultimately increasing the solar cell recoil to 99%. Therefore, if you put a cheaper PWM controller, then increase the array of solar cells by another 20%.
Calculation of solar panels for a private house or cottage
If you do not know your consumption and only plan to say that you are supplying a dacha from solar panels, then consumption is considered quite simple. For example, you will have a fridge working at your dacha, which consumes 370 kWh per year, which means that in a month it will consume only 30.8 kWh of energy, and on the day of 1.02kWh. Also light, for example, light bulbs you have energy-saving say for 12 watts each, there are 5 of them and they shine an average of 5 hours a day. This means that per day your light will consume 12 * 5 * 5 = 300 watts of energy, and for a month "burn" 9kW * h. You can also read the consumption of the pump, TV and everything else that you have, add up everything and get your daily energy consumption, and multiply it by a month and get some approximate figure.For example, you get 70kWh of energy per month, add 40% of energy, which will be lost in the battery, inverter, etc. So we need the solar panels to produce about 100kW * h. This means 100: 30: 7 = 0.476kW. It turns out you need an array of batteries with a power of 0.5 kW. But such an array of batteries will suffice only in the summer, even in the spring and autumn with cloudy days there will be interruptions with electricity, so you have to increase the array of batteries in half.
As a result of the foregoing, in brief, the calculation of the number of solar batteries looks like this:
Example: Consumption of private house 300kWh per month, we divide by 30 days = 7kW, we divide 10kW for 7 hours, we will get 1.42kW. Add to this figure 40% of the losses on the battery and in the inverter, 1.42 + 0.568 = 1988 watts. As a result, to power a private house in the summer time you need an array of 2 kW. But in order to get enough energy even in spring and autumn, it's better to increase the array by 50%, that is, plus 1kW. And in winter, in prolonged overcast periods, use either a gasoline generator, or install a wind generator with a power of at least 2 kW. More specifically, you can calculate based on the weather archive data for the region.
Cost of solar batteries and batteries
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Prices for solar panels and equipment are now quite different, one and also the products can differ in price from different sellers, so look for cheaper and time-tested sellers. Prices for solar panels now average 70 rubles per watt, that is, an array of batteries in 1 kW will cost about 70,000 rubles, but the more the lot, the more discounts and cheaper shipping.
Qualitative specialized batteries are expensive, a battery of 12v 200Ah will cost an average of 15-20 rubles. I use this kind of acb, I write about them in this article Accumulators for solar batteries Automobiles are two times cheaper, but they should be put twice as much so that they have served at least five years. And also car batteries can not be installed in residential premises, as they are not airtight. Specialized in the discharge not bllee 50% will serve 6-10 years, and they are sealed, do not distinguish anything. You can buy and cheaper if you take a large batch, usually sellers give decent discounts.
The rest of the equipment is likely to be individual, inverters are different, and in power, and in the shape of a sinusoid, and at a price. Also, the charge controllers can be as expensive with all the functions, including with connection to the PC and remote access via the Internet.
Solar panels are becoming an increasingly demanded type of autonomous systems every year, which are an alternative to traditional power supply. Especially popular are the installation of solar collectors in the suburban zone, in dachas where there is no supply of electricity.
Power calculations
When buying solar panels for the home, the owners, first of all, are interested in how much power the batteries will need in order to satisfy all the vital needs. Since the system can provide electricity to many devices only when the energy consumption is not higher than the amount of energy produced by the generator.
The system consists of the main 4 components:
- Accumulators;
- The charge controller;
- Photovoltaic panels;
- Inverter.
Calculating the power of solar panels for the house is relevant, above all, in that, with all financial and material constraints, it is important to know what kind of result to expect from batteries and whether it is worth buying such a power supply system at all. For each mode of electricity use, there is a system of calculation.
The basis for calculating the required energy is taken from the data on the possibility of the Sun (the power of solar radiation), and also it is necessary to calculate how much energy is planned to consume. This can be done independently by looking at the table "Calculation of electricity consumption":
This takes into account:
- Region;
- Weather;
- The angle of the panel.
When setting the angle of the panel, it is important to determine whether the batteries will be used year-round or only in summer. Preferably, the angle of inclination for panels is 15 ° greater than the geographical latitude. The more the slope, the more efficient the production of energy.
Calculation of solar panels for the house is desirable to carry out, having the data for both horizontal and vertical panel settings.
Calculation process
In order to evaluate the performance of solar panels, it is desirable to take for the calculations the worst month in winter (January in Moscow) and the summer high (July in Moscow).
Standard stream of sunlight at 25 ° in 1 kW / m? Is the nominal power of the solar panel. Taking a monthly insolation, and multiplying it by the ratio of the maximum insolation and battery power, you can get an estimate of the battery output for a particular month.
The production of photovoltaic panels is calculated by the formulas:
1. Eb = Einx x Pcb x? / Pince
Esb - energy by a solar battery;
Eins - Insolation of 1 m? (the specific month from the table);
?
- Efficiency of electric current transmission;
Pbb - rated power of the battery;
Pince - maximum insolation power 1 m? of the earth's surface.
You can also do the calculation of the power of solar cells needed for monthly energy consumption.
2. Рсб = Ринс х Ебб / (Еинс х?)
In calculating the efficiency, it is possible to lay the losses (from 10 to 25%) that can occur from cheap charge controllers, which, as a rule, either underestimate the output voltage of the battery or simply ignore surplus energy.
2 The formula is convenient if it is necessary to calculate the rated power of solar cells, taking into account the specific conditions of insolation, but it is not very suitable for calculating the possibilities for the whole year.
1 The formula allows you to calculate the power for different modes of power supply of batteries with different nominal power.
Example calculation for Moscow
Suppose that you need to calculate the slope of 70 °, but for Moscow there is no such data, but there are data on the angle of inclination of the panel 40 ° and 90 °. In this case, the average value is taken between these data and rounded up to 1 kWh by a smaller number. When calculating the power, the total efficiency of the controller and inverter is taken into account - 91%. The value of "deficit mode" indicates that the capacity is not enough even for the permanent operation of the system itself.
Analysis of calculations
Given the weather characteristics and the rated power of the batteries, it can be concluded that 400 W of the battery in Moscow will not be enough even to support the emergency mode in the summer. Although 80% of the emergency level can be considered an acceptable option, especially when the inverter operates erratically, but only when power is needed.
Low-power systems are not designed for round-the-clock household power supply even in summer. Since the energy in such systems is critical for own consumption of the charge controller and inverter. In winter, the power solar collector will not be enough for the operation of all electrical appliances at home, but in summer it is entirely permissible that electricity supply will be uninterrupted.
Battery capacities from power calculations for Moscow:
- 500 W - gives an emergency minimum of 80% from May to the end of August;
- 600 W - the middle of March - September;
- 800 W - with excess of emergency level (except December and January) provides voltage from March to September;
- 1 kW - provides basic power consumption almost all year round, but in winter (December and January), energy may not be enough;
- 1.2 kW - provides a moderate mode in July, in March - September, the energy consumption mode is basic. The emergency minimum falls on the period November - January;
- 2 kW - maintains a comfortable mode, or close to it in the period May - August and base from February to August. But in the long dark months of a given capacity of the solar collector may not be enough;
- 3.2 kW - provides a comfortable mode for all long days and during the year allows you to count on an emergency minimum;
- 5.3 kW - batteries of rated power, which allow to use electric power practically without restrictions in the period May - August and all year round in the basic mode;
- 8 kW - the power of the solar battery, which ensures the use of electricity all year round in a moderate mode;
- 13.5 kW - year-round comfortable mode of energy consumption.
Basic selection criteria for equipment
The provision of electricity from solar collectors is affected by:
- Duration of the day and night (at night solar systems stop feeding energy);
- Weather conditions (on cloudy days, the level of energy supply decreases);
- Seasons (when the days become shorter than nights).
- Only for the summer period - not less than 400 A / h per 1 kW / h of daily consumption in the minimum regime;
- For year-round energy consumption - not less than 800 A / h per 1 kW / h in the minimum consumption mode.
When choosing a panel, three main factors are taken into account:
- Geometry;
- Type of photocells;
- Rated output voltage.
When there is a question: "to acquire one large panel or a few small ones", our advice is better one. Small panels are good to install where there is no possibility to install a large panel (its size does not exceed 1.5 - 2 meters). In this case, the area of connections will be smaller, and the level of reliability will increase.
The most common types of photocells are:
- On single-crystal silicon;
- On polycrystalline silicon.
Monocrystalline type is more expensive, but its advantages are much higher than polycrystalline type.
If the total power of the panels exceeds the power of the inverter, this will be justified several times even taking into account the constant powerful load and a powerful battery pack.
When choosing the placement of panels, the orientation of the house around the world and its "landing" on the terrain are taken into account. The traditional orientation is the placement of panels to the south.
Now there is no problem to acquire a system for tracking the sun. Will be justified the cost of such additional equipment for the solar collector or not - the decision is purely individual.
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Calculation of a solar battery for a house or cottage
To understand how many solar cells are needed and how much power they will generate, you need to know how much energy is needed to provide all the consumers located in the building.
A little about the equipment
For full-fledged operation of the home energy complex, it is necessary to use the following set of equipment:
- (inverter).
Calculating the power of solar panels
Calculation of the solar battery for the house must begin with the calculation of the need for electric power. This can be solved in two ways. You can analyze the meter, and you can calculate the amount of installed capacity of all consumers. This list includes:
- heating equipment;
- fridge;
- washer;
- lighting, etc.
For convenience, we present a table of average household electricity consumption
| Consumer | Power | Season | Duration of work per day | Daily Consumption | ||
| average | maximum | Average | maximum | |||
| Main regular consumers | ||||||
| Inverter | 20 W | always | 24 hours | 1.73 MJ (0.48 kWh) | ||
| Charge Controller | 5 W | always | 24 hours | 0.43 MJ (0.12 kWh) | ||
| Lighting | 200W | winter | 8 ocloc'k | 10 hours | 5.76 MJ (1.6 kWh) | 7.2 MJ (2 kWh) |
| (simultaneously 10 energy-saving pampas of 20 W. similar incandescent bulbs of 100 W) | summer | 2 hours | 4 hours | 1.44 MJ (0.4 kWh) | 2.88 MJ (0.8 kWh) | |
| Fridge | 500W | winter | 2 hours | 2.5 hours | 3.6 MJ (1 kWh) | 4.5 MJ (1.25 kWh) |
| (compressor operation) | peto | 2.5 hours | 3 hours | 4.5 MJ (1.25 kWh) | 5.4 MJ (1.5 kWh) | |
| Vibration pump | 250 W | winter | 30 minutes | 40 minutes | 0.45 MJ (0.125 kWh) | 0.6 MJ (0.17 kWh) |
| summer | 2 hours | 3 hours | 1.8 MJ (0.5 kWh) | 2.7 MJ (0.75 kWh) | ||
| The centrifugal pump | 800 W | always | 15 minutes | 30 min | 0.72 MJ (0.2 kWh) | 1.44 MJ (0.4 kWh) |
| Washing machine (mechanical washing and spinning, but without heating the water) | 500W | always | 1 hour | 6 hours | 1.8 MJ (0.5 kWh) | 10.4 MJ (3 kWh) |
| Iron (taking into account the work of the thermostat) | 1500 W | always | 30 minutes | 2 hours | 2.7 MJ (0.75 kWh) | 10.4 MJ (3 kWh) |
| TV with video player or VCR | 150 W | always | 2 hours | 4 hours | 1.08 MJ (0.3 kWh) | 2.16 MJ (0.6 kWh) |
| A laptop | 100 W | always | 2 hours | 4 hours | 0.72 MJ (0.2 kWh) | 1.44 MJ (0.4 kWh) |
| TOTAL | up to 2.5 kW maximum, usually not more than 1.5 kW | winter | 19.5 MJ (5.5 kWh) | 41 MJ (11.5 kWh) | ||
| summer | 15 MJ (4.5 kWh) | 39.5 MJ (11 kWh) | ||||
| Secondary regular consumers | ||||||
| Electric kettle | 2 kW | always | 5 times for 3 minutes | 20 times for 3 minutes | 0.9 MJ (0.5 kWh) | 7.2 MJ (2 kWh) |
| Kitchen water heater | 1.2 kW | winter (from 5 С) |
2 hours (25 liters) |
5 o'clock (50 liters) |
9 MJ (2.5 kWh) | 19.5 MJ (5.5 kWh) |
| heating water to 70 ° C. heated batch of 10 liters | peto (from 15C) | 1.5 hours (25 liters) | 3 hours (50 liters) | 5.5 MJ (1.5 kWh) | 11.5 MJ (3.2 kWh) | |
| Hot Water Boiler | 0.7 / 1.3 / 2.0 kW | winter (with 5С) | 4/2 / 1.25 hours (50 liters) | 12/6/4 hours (150 liters) | 9.5 MJ (2.5 kWh) | 28 MJ (8 kWh) |
| heating water for bath and shower up to 50еС, heated portion is not more than 100 liters | peto (from 15C) | 3 / 1.5 / 1 hour (50 liters) | 10/5/3 hours (150 liters) | 7 MJ (2 kWh) | 21.5 MJ (6 kWh) | |
| TOTAL | up to 4 kW maximum, usually no more than 2 kW | winter | 20 MJ (5.5 kWh) | 56 MJ (15.5 kWh) | ||
| summer | 14.5 MJ (4 kWh) | 41 MJ (11.5 kWh) | ||||
| Irregular consumers | ||||||
| Kitchen electrical appliances (food processor, meat grinder, mixer, juicer, etc.) | up to 2 kW | always | 30 minutes | 4 hours | up to 1.8 MJ (1 kWh) | up to 14.4 MJ (4 kWh) |
| Cosmetic electrical appliances (electric razor, hair dryer, etc.) | up to 2 kW | always | 5 minutes | 30 minutes | up to 0.3 MJ (0.15 kWh) | up to 1.8 MJ (1 kWh) |
| A vacuum cleaner | 1800 W | always | 30 minutes | 2 hours | 3.5 MJ (0.9 kWh) | 13 MJ (3.6 kWh) |
| The electrotool (a Bulgarian, a drill, a jigsaw, electric saws, etc.) |
up to 2 kW | always | 1 hour | 4 hours | up to 3.6 MJ (1 kWh) | up to 14.4 MJ (4 kWh) |
| Lawn mower or trimmer | 1500 W | peto | 1 hour | 4 hours | 5.4 MJ (1.5 kWh) | 18 MJ (5 kWh) |
| Snowplower | 1500 W | winter | 1 hour | 4 hours | 5.4 MJ (1.5 kWh) | 18 MJ (5 kWh) |
| TOTAL | up to 2 kW | |||||
Suppose the total consumption is 100 kWh in one month, then it means that the solar panels should produce just as much electricity.
Solar panels , installed in the yard or on the roof, are able to produce energy only if available. Maximum (passport) power they reach with a cloudless firmament and light reaching their surface at an angle of 90 degrees. At other angles, the energy produced is significantly reduced. Moreover, in cloudy weather, it can fall 15-20 times. All this is necessary to know by performing the calculation of solar panels for a private house.
When calculating the number of solar cells for a house, it makes sense to focus on working hours, it is during these hours that solar panels work at full capacity. In the morning and evening time, the amount of generated energy will be from 20 to 30% of the installed capacity, and the rest will be generated during working hours.
Calculating the power of solar panels for the house shows - a panel with a capacity of 1 kW per summer day, it is guaranteed to produce 7 kW per day or 210 kW per month. You can, of course, add the amount that will be produced in the twilight time of the day (morning and evening), but it is better to consider it a reserve in case of changing weather conditions. By the way, if the panels are installed in one place, then, of course, they will not generate all the specified capacity. That is, if the homeowner installs a panel with a total capacity of 2 kW, then a month it will generate about 420 kW of energy. Also, the number of kilowatts produced depends on the level in your region.
What to consider when calculating solar panels for the home
How to calculate the power of solar panels for a home with a loss? Of course, to have a volume of electricity in 420 kW per month is not bad, but we must bear in mind that in our country there is no completely sunny months. Certainly it turns out that a few days will be overcast, that is, from the resulting figures, you can safely cross out those days. Accordingly, at the disposal of the homeowner will not 420 kW, but somewhat less, for example, 360.
In addition, it is necessary to understand that in the off-season a light day is less, and cloudy days are longer. That is, if you use the energy of the sun from March to October, it makes sense to increase the number of solar cells by 30-50%, but this depends on the specific area. To get electricity in winter, you can forget because of a short bright day and a lot of clouds in the sky.
In addition, the above should take into account the losses that are inevitable in the batteries and the converter.
Loss on battery and inverter
The necessary amount of energy in the dark should be enough to survive. If you consume 3 kWh, the batteries should contain exactly this amount of energy. But, they can not be completely discharged, for example, car batteries, you can empty 50%. You can calculate the estimated stock of stored energy - with a daily consumption of 10 kWh, the capacity of the battery should be equal to this figure.
Inverters, which are an integral part of the solar energy system, have an efficiency factor of 70-80%.
Thus, it can be concluded that from the use of battery, inverter, controller, the system will lose from 40 to 50% of the generated power. That is, the landlord will need to increase the number of panels for these lost interest and this figure can change the calculation of the cost of solar panels for a private house.
Rules for calculating the number of solar cells for the house
To calculate the solar system power, you can use the following rules:
- determine that the most effective solar panels work only hours per day;
- perform a calculation of energy consumption, divide the result by 7 and the required power of solar panels;
- add to the result 40 - 50% percent, to compensate for losses from the battery and inverter.
The use of energy received from the sun is good, but using it as the main thing is probably not entirely appropriate, especially in our climatic conditions.
