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Home » Solar inverters » 1kW solar panel produced electricity on a sunny day
1kW solar panel produced electricity on a sunny day
Q:
How much amount of electricity a 1 kw Solar Panel can produce per day on a sunny day? (of eight hours)
A:
It depends on the solar radiation on the location at the time of the year. Notably, the latitude of the location and there is a way to calculate the available solar radiation for the sunny clear day. But the space is limited here so I use measured average solar radiation (not the sunny day) in December (because we use the lowest radiation in the year for estimation).
For Montreal where I live, the electric energy produced by the panel of 1 kW per square meter becomes roughly 1.8 kWh per day by rough calculation. And it would be more for the rest of the year, of course.
The calculation I followed is:
Average daily solar radiation on the horizontal surface for December is 1.580 kWh per square meter per day. If the panel is installed in the angle of the latitude, average daily solar radiation on the panel surface becomes 2.253 kWh per square meter per day. Taking into account the losses, it becomes 1.825 kWh per square meter per day. The power of the panel at the Standard Testing Condition is 1 kW (assuming per 1 square meter) as you specified. This gives, so called, Peak Solar Hours of 1.8 hour per day. Assuming Peak Power is 1 kW, this gives 1.8 kWh of generated energy.
How much amount of electricity a 1 kw Solar Panel can produce per day on a sunny day? (of eight hours)
A:
It depends on the solar radiation on the location at the time of the year. Notably, the latitude of the location and there is a way to calculate the available solar radiation for the sunny clear day. But the space is limited here so I use measured average solar radiation (not the sunny day) in December (because we use the lowest radiation in the year for estimation).
For Montreal where I live, the electric energy produced by the panel of 1 kW per square meter becomes roughly 1.8 kWh per day by rough calculation. And it would be more for the rest of the year, of course.
The calculation I followed is:
Average daily solar radiation on the horizontal surface for December is 1.580 kWh per square meter per day. If the panel is installed in the angle of the latitude, average daily solar radiation on the panel surface becomes 2.253 kWh per square meter per day. Taking into account the losses, it becomes 1.825 kWh per square meter per day. The power of the panel at the Standard Testing Condition is 1 kW (assuming per 1 square meter) as you specified. This gives, so called, Peak Solar Hours of 1.8 hour per day. Assuming Peak Power is 1 kW, this gives 1.8 kWh of generated energy.
How can we apply these figures in Egypt?
The latitude of Egypt is lower than Montreal, Canada. And I'm assuming it's dry even in coastal area, meaning less cloudy. So availability of solar radiation would be higher. And I assume the seasonal variation would be lower. While the performance of solar panels decreases as the temperature rises, more availability of solar radiation would have more impact. So in general, there would be more electric energy generated.
About the figures, first of all, it's really a rough calculation. And we need information about how much power per square meter for a particular panel, i.e. how much W [Watt] / m^2 [square meter]. I simplified the calculation only taking its essence.
In the case of Egypt, I assume that the energy consumption is highest in the hottest month for air conditioning, while here in Montreal it's in winter for heating (, which why I used the average daily solar radiation in December). At the same time, I assume availability of solar radiation is highest in that month in Egypt. So I would use the 12 months average of the average daily solar radiation for calculation.
Unfortunately, I don't have a time to do the calculation for Egypt. But as a reference, the same calculation I did before for Dubai in UAE (hot dry climate) using 12 months average solar radiation gave very roughly 6.4 kWh (without considering performance degradation by higher temperature).
About the figures, first of all, it's really a rough calculation. And we need information about how much power per square meter for a particular panel, i.e. how much W [Watt] / m^2 [square meter]. I simplified the calculation only taking its essence.
In the case of Egypt, I assume that the energy consumption is highest in the hottest month for air conditioning, while here in Montreal it's in winter for heating (, which why I used the average daily solar radiation in December). At the same time, I assume availability of solar radiation is highest in that month in Egypt. So I would use the 12 months average of the average daily solar radiation for calculation.
Unfortunately, I don't have a time to do the calculation for Egypt. But as a reference, the same calculation I did before for Dubai in UAE (hot dry climate) using 12 months average solar radiation gave very roughly 6.4 kWh (without considering performance degradation by higher temperature).
Egypt is close to the same sunhours as here in the sunny southwest of the USA. Our annual insolation hours here is 6.5 hours (winter min of 5 hrs) so a simple daily production of the system can be figured out with a simple single derate factor (see link below for ways to get this). Temperature is very important but i try to capture that in the derate as well.
1. kw (DC) x 6.5 sunhours/day x .8 (DC to AC derate factor) = 5.2 kw hrs/day
Yearly = 5.2 kw hrs/day x 365 days/year = 1,898 kw hrs/year
1. kw (DC) x 6.5 sunhours/day x .8 (DC to AC derate factor) = 5.2 kw hrs/day
Yearly = 5.2 kw hrs/day x 365 days/year = 1,898 kw hrs/year
A number of factors must be taken into account when attempting to calculate PV system yield based on known irradiance values. First and foremost, the efficiency of the solar modules. Lets go with basic calculation:
The total surface area of your modules adds up to 7m2 with the overall capacity of 1kw. Under STC (1000W/m2 Standard Test Conditions) an output of 1kw is produced from 7kw of irradiance, which is equivalent to an efficiency if 14.2 percent. So the calculation for the chosen month (with xx kwh/m2) should look like this: 7m2 (total module area) times xx kwh/m2 times the efficiency. This is the theoretical way to calculate. In real conditions you have the losses and possible gain by south facing alignment the modules.
The total surface area of your modules adds up to 7m2 with the overall capacity of 1kw. Under STC (1000W/m2 Standard Test Conditions) an output of 1kw is produced from 7kw of irradiance, which is equivalent to an efficiency if 14.2 percent. So the calculation for the chosen month (with xx kwh/m2) should look like this: 7m2 (total module area) times xx kwh/m2 times the efficiency. This is the theoretical way to calculate. In real conditions you have the losses and possible gain by south facing alignment the modules.
When we design a solar system, there would be several parts we should consider. One is the solar radiation, therefore basically we would ask our customer which country,city the system would be placed. Second would be the place to install these panels, like the angle of the panels, on the ground or the roof and so on. Third would be the energy losses, like the inverter efficiency, cable losses, distance between the solar panels and inverter, battery and other facilities. Besides, the shade, temperature, humidity etc also effects...Therefore, it is kind a complicated project.
I realized that I missed many things and relearned a lot by reviewing what I had learned before.
Not necessarily related to that issue but, my very rough calculation is based on the concept of Peak Solar Hours. In short, instead of using the actual hours of sunlight with varying strength (power), it's using the shorter hours of the equal amount of solar power. e.g. Daily solar radiation of 1.8 kWh for 8 hours with varying strength (power) throughout the day (starting from 0 kW in the morning, highest around noon, and ending with 0 kW in the evening) equals the daily solar radiation of 1.8 kWh for 1.8 hours with the constant 1 kW (power).
Not necessarily related to that issue but, my very rough calculation is based on the concept of Peak Solar Hours. In short, instead of using the actual hours of sunlight with varying strength (power), it's using the shorter hours of the equal amount of solar power. e.g. Daily solar radiation of 1.8 kWh for 8 hours with varying strength (power) throughout the day (starting from 0 kW in the morning, highest around noon, and ending with 0 kW in the evening) equals the daily solar radiation of 1.8 kWh for 1.8 hours with the constant 1 kW (power).
1 kw Solar panel
Theoretically : 1Kw x hrs of sunshine(various for different areas)
Eg for Chennai : 1KW x 5 hrs = 5Kwhr
But the 1 Kw solar panel gives the power under standard test condition.like temperature.AM,insolation etc.
So practically it will give around 3 kwhr per day (during a clear sunny day)
By using MPPT .it can be increased to 3.6 Kwhr
these data is for chennai
Theoretically : 1Kw x hrs of sunshine(various for different areas)
Eg for Chennai : 1KW x 5 hrs = 5Kwhr
But the 1 Kw solar panel gives the power under standard test condition.like temperature.AM,insolation etc.
So practically it will give around 3 kwhr per day (during a clear sunny day)
By using MPPT .it can be increased to 3.6 Kwhr
these data is for chennai
It totally depends on where you are, and how much light falls on the panel. All I can tell you that in Adelaide the average daily output (without shade, and facing North) is about 4.2 kWh (less derating, inverter efficiency and design efficiency). Don't forget to take away the number you first thought of :)