Summary: With the deepening of the concept of energy saving and emission reduction, photovoltaic power generation has developed rapidly in recent years. Today, many ordinary people have installed photovoltaic solar power generation in their homes. However, some users who are considering installation do not know how photovoltaic power generation is charged, how much photovoltaic power generation costs per square meter, or whether to install monocrystalline photovoltaic power generation panels or polycrystalline photovoltaic energy production panels. The editor below will answer these questions. 1. How much does photovoltaic power production cost per square meter?
The price of photovoltaic solar cell modules is usually based on power (watts) rather than square meters of area. The current market price of a setfull photovoltaic power generation is about 8 yuan/W. The power of each photovoltaic power generation module commonly used in China is 270 W. It can be calculated that the price of each module is 270*8 = 2160. yuan.
General situation: The installation area of 3KW system is about 22_, 4KW system is about 30_, 5KW system is about 40_, 6KW system is about 50_ and 1kw10 square meters.
2. Are monocrystalline or polycrystalline photovoltaic solar panels better?
Photovoltaic solar modules currently on the market can be roughly divided into three categories: monocrystalline silicon modules and polycrystalline silicon modules. and amorphous silicon modules (thin film modules).
1. Monocrystalline silicon components
The current photoelectric conversion efficiency of monocrystalline silicon solar cells is enaround 15%, the highest reaching 24%. It has the highest photoelectric conversion efficiency, but its production cost is so high that it cannot be widely and commonly used in large quantities. Since monocrystalline silicon is typically encapsulated in tempered glass and waterproof resin, it is strong and durable, with a lifespan of typically up to 15 years and up to 25 years. It is currently the dominant trend in the market.
2. Polycrystalline Silicon Components
The manufacturing process of polycrystalline silicon solar cells is similar to that of monocrystalline silicon solar cells, but the photoelectric conversion efficiency of polycrystalline silicon solar cells is much lower. . About 12%. In terms of production cost, it is cheaper than monocrystalline silicon solar cells. The material is easy to manufacture, saves energyogy and the overall production cost is low, so it has been widely developed. In addition, the lifespan of polycrystalline silicon solar cells is shorter than that of monocrystalline silicon solar cells. In terms of performance-price ratio, monocrystalline silicon solar cells are slightly better.
3. Amorphous Silicon Components (Thin Film Components)
Even if scrap silicon wafers are used, silicon wafers are not necessarily inexpensive given their efficiency levels. Thin-film solar cells are cheaper than traditional solar panels, but are also less efficient, with photovoltaic conversion rates between 20% and 30%. Compared with monocrystalline silicon components and polycrystalline silicon components, they are basically 2 to 3 generations behind.
For users, it doesn't matter whetherthey choose monocrystalline, polycrystalline or thin film modules, the most important thing is to choose the overall income! Product quality is the most important factor in determining a power plant's revenue. If your product is of good quality, you will earn more. Regardless of the type of component it is, the same principle applies.
How to calculate the number of kilowatt-hours of electricity generated by photovoltaic energy production per day
Question 1: How to calculate the number of kilowatt-hours of electricity generated by photovoltaic energy production per day ? A 1 kW module has effective sunshine for 6 hours, and does not take into account losses, i.e. 6 kilowatt hours of electricity per day? . The loss of independent systems is generally 30%. 6*0.7=4.2 kW/h. You must take into account the intensity of solar radiation. 6 hours of effective sunshine refers to the intensity of sunshineent. The so-called effective hours of sunshine refer to the intensity of the radiation.
Example:
If the unit of radiation is cal/cm2, then:
Maximum sunshine hours = radiation × 0.0116
0.0116 is the conversion factor to convert radiation (cal/cm2) to maximum sunshine hours:
Definition of maximum sunshine: 100 mW/cm2=0.1 W /cm2
1cal=4.1868 J=101868Ws 1h =3600s
Then: 1cal/cm2=4.1868Ws/cal(3600s/h×0.1W/cm2)=0 .0116hcm2/cal
For example: Suppose a certain place The annual horizontal radiation is 135 kcal/cm2 and the radiation on the square surface is 148.5 kcal/cm2, then the hours d The maximum annual sunshine hours are:
148500×0.0116=1722.6 h, the maximum daily sunshine hours are
p>
1722.6÷365=4.7h
Question 2: Concerning the calculation of the annual production of photovoltaic energy, an estimateAn approximation can be made as follows: Total electricity production (kWh) = effective surface area of photovoltaic cells (square meters) * annual average intensity of total solar radiation (w/square meters) * annual effective sunshine duration (hours) * component efficiency (silicon cells are usually 15%) * system efficiency ( Usually 75% is acceptable.) If you want to calculate carefully, you need to use special software.
Question 3: How is photovoltaic energy production calculated? Photovoltaic generator machine capacity * equivalent sunshine time = initial total energy production.
Total initial energy production * panel efficiency = panel energy production
The unit of equivalent sunshine time is hour/day. Therefore, the calculated electricity production of the square network is the electricity productionaverage daily tricity. To calculate the annual electricity production, it must be multiplied by 365.
The equivalent sunshine duration can be obtained using "the average annual amount of solar radiation" / 1000W.
Square network efficiency = (1-cable loss)*equipment efficiency*transformer efficiency*(1-dust impact)
< p> The efficiency of square array is generally 70%~80%, maybe 70% is not that low, I don't remember exactly. How far should you go? You can apply to a design institute with new energy qualifications.
Question 4: How to calculate the actual energy production of photovoltaic modules? Actual power of photovoltaic modules * effective lighting duration * total system efficiency (usually refers to efficiency loss subtracted) = actual energy production < /p>
Question 5: How is photovoltaic energy production calculated? 100 square meters can be installed with battery panels of around 15 kilowatts, and the inverter is selected according to the capacity of the battery panel.
Average electricity production is about 60 degrees per day!
Question 6: How to calculate the electricity production of a photovoltaic plant? The annual electricity production of a photovoltaic plant has a great relationship with the capacity of the system installed in the power supply? station and local average annual sunshine hours. The specific calculation The formula is as follows for reference: annual electricity generation = system capacity * average annual sunshine hours * 0.8 * 365; (This answer is referenced by solar energy expert Guangdong Sunku New Energy)
Question 7: How to calculate the productiondaily energy ction of solar energy production To estimate the energy production of. the photovoltaic power generation system, you need to know the system installation Local effective sunshine hours, system efficiency and system installation capacity.
For example, for a 1000W grid-connected photovoltaic system, the installation location is Beijing, the effective sunshine duration is 4 hours, and the efficiency of the grid-connected photovoltaic system is is about 80%, so the daily power system production calculation formula = effective module installation capacity Hours of sunshine system efficiency = 1,000 × 4 × 0.8 = 3,200 Wh, or about 3 .2 kilowatt hours of electricity.
Question 8: How is the electricity production of a photovoltaic plant calculated? 5 points Are you referring to how the rated power of a photovo plantltaic is calculated?
Capacity of the photovoltaic field * equivalent duration of sunshine = total initial electricity production.
Total initial energy production * panel efficiency = panel energy production
The unit of equivalent sunshine time is hour/day. Therefore, the calculated electricity production of the square grid is the daily average electricity production. To calculate the annual electricity production, it must be multiplied by 365.
The equivalent sunshine duration can be obtained using "the average annual amount of solar radiation" / 1000W.
Square network efficiency = (1-cable loss)*equipment efficiency*transformer efficiency*(1-dust impact)
< p> The efficiency of square array is generally 70%~80%, maybe 70% is not that low, I don't remember exactly. How far should you go?? You can apply to a design institute with new energy qualifications.
Question 9: Calculate the energy production of solar panels and the panel area required for a 1 MW rooftop photovoltaic plant.
The area of a 235W polycrystalline solar panel panel is 1.65*0.992=1.6368O, 1MW requires 1000000/235=4255.32 batteries, the total area of the battery panel 1.6368*4255.32= 6965O
Theoretical annual energy production = annual average total solar radiation * total battery area * photoelectric conversion efficiency = 5555.339*6965*17.5% < /p>< p > =6771263.8MJ=6771263.8*0.28KWH=1895953.86KWH
=1.896 million degrees
Actual energy production efficiency< /p>
The DC power produced by the solar panel is the Rated power of the panel. Solar panels operating on site often do not meet the requirementsstandard test tions and the allowable output deviation is 5%. Therefore, the influence coefficient of 0.95 should be considered when analyzing the power output of solar panels.
As the temperature of the photovoltaic module increases, the output power of the f:l group will decrease. For crystalline silicon modules, when the temperature inside the photovoltaic module reaches 50-75°C, its output power drops to 89% of the rated value. The impact of 0.89 should be considered when analyzing the power output of the solar panel. coefficient.
The accumulation of dust on the surface of photovoltaic modules will affect the intensity of solar radiation radiated to the surface of the panel, and will also affect the output power of the solar panel. According to relevant literature reports, this factor will have an impact of 7% on the production of photovoltaic modules. During the analysise of the output power of the solar panels, an impact coefficient of 0.93 must be taken into account.
Due to the inhomogeneity of solar radiation, it is almost impossible for the output power of photovoltaic modules to reach the maximum power at the same time, so the output power of the photovoltaic generator is less than the sum of the nominal powers of each module.
In addition, there is a mismatch of photovoltaic modules and loss of panel connection, etc. The coefficient of these factors affecting the output power of solar panels is calculated to be 0.95.
The efficiency of the grid-connected photovoltaic plant after taking into account the installation angle factor is 0.88.
The real efficiency of electricity production is therefore O. 9 5 * 0.8 9 * 0.9 3*0.9 5 X*0.8 8=6 5.7% .
, The actual annual electricity production of the production systemn of photovoltaic energy = Theoretical annual electricity production * Actual electricity production efficiency
=189.6*0.9 5 * 0.8 9 *0.9 3*
0.9 5 * 0.8 8=189.6*6 5.7%=1.2456 million degrees
This varies from company to company and is for reference only
< p> Question 10: How to calculate the number of kilowatt hours of photovoltaic energy generated per day for a 1KW module with 6 hours of effective sunshine, this makes 6 kilowatt hours of electricity per day. The loss of independent systems is generally 30%. 6*0.7=4.2 kW/h. You must take into account the intensity of solar radiation. 6 hours of effective sunshine refers to the intensity of sunshine. The so-called effective hours of sunshineThe number refers to the intensity of the radiation.
Example:
If the unit of radiation is cal/cm2, then:
Maximum sunshine hours = radiation × 0.0116
0.0116 is the conversion factor to convert radiation (cal/cm2) to maximum sunshine hours:
Definition of maximum sunshine: 100 mW/cm2=0 .1 W/cm2
1cal=4.1868 J=101868Ws 1h =3600s
Then: 1cal/cm2=4.1868Ws/cal(3600s/h×0.1W/cm2 )=0.0116hcm2/cal
For example: Suppose a certain place The annual horizontal radiation is 135 kcal/cm2 and the radiation on the square surface is 148.5 kcal/cm2, then the annual maximum sunshine hours are:
148500×0.0116=1722.6 h, the daily maximum sunshine hours are
p>
1722.6÷365 =4.7h
1. The 1KW module has effective sunshine for 6 hours and generates 6 kWh of electricity per day without taking into account losses. The loss of independent systems is generally 30%.
2. Taking into account the intensity of solar radiation, 6 hours of effective sunshine correspond to 6*0.7=4.2 kW/h. It produces 4,2 kWh of electricity per day. Effective daylight hours refer to the intensity of the radiation.
3. Daily solar power output = sunshine duration * total photovoltaic field power * power generation efficiency
4. The main principle of photovoltaic power generation is the photoelectric effect of semiconductors. When a photon irradiates a metal, all of its energy can be absorbed by an electron in the metal. The energy absorbed by the electron is large enough to overcome the internal gravity of the metal to do work, escape from the metal surface and become. photoelectrons.
Detailed information:
1. The kilowatt hour is what we usually call the “degree”. ", is the unit of electric power, symbol: kWh, and the calculation formula is power multiplied by time. Suppose the power of un energy-consuming device is 2,500 watts, that is to say its electricity consumption per hour is 2.5 kilowatt hours, or 2.5 kilowatt hours of electricity per hour.
The units of work include joules and kilowatt hours, and the relationship between them is:
1 joule = 1 watt × second
1; kilowatt hour = 1 KW =1kW×h=1000W×h=1000W×3600s=3600000J;
For daily use, 1 kilowatt hour is equivalent to 1 degree.
2. Efficiency Attenuation
After the crystalline silicon photovoltaic modules are installed and exposed to sunlight for 50-100 days, the efficiency will attenuate by about 2-3%. significantly and stabilizes at an annual attenuation of 0.5 to 0.8% for 20 years. The attenuation is approximately 20%. Monocrystalline components have less attenuation than polycrystalline components. The attenuation of amorphous optical components is approximately lower than that of silicon ccrystalline.
Improving the conversion rate and reducing the cost per watt will therefore remain the two major themes for the future development of photovoltaics. Regardless of the method used, if large-scale application can increase the conversion rate to 30% and the cost is less than 5,000 yuan per kilowatt (the same as hydropower), then humanity will have the largest and cleanest energy production by nuclear fusion. before the search succeeds, the cheapest and almost unlimited new reliable energy.
References:
Baidu Encyclopedia-Photovoltaic Power GenerationReferences: < / p>Baidu Encyclopedia Diploma