When trying to decide if rooftop solar is right for you, the amount of electricity produced is perhaps the most significant variable to consider. The amount of electricity produced is dependent on a series of factors, including the amount of sunlight, the size of the solar panel array, and the efficiency of your PV system in converting sunlight into DC and then AC electricity.
Your solar provider will undoubtedly be able to provide you with a detailed breakdown of the expected system performance after a brief consultation. However, if you’re the type that needs a deeper understanding of the numbers, you’ve come to the right place. In this post, we will walk you through each of these aspects, allowing you to make an informed decision when it comes to deciding on a solar system.
The amount of sunlight
It’s relatively easy to comprehend why the amount of sunlight your roof sees over the year is a significant factor in PV system performance. However, it might not be as clear how to quantify this or find the data for your location.
First, the technical term to get familiar with is irradiation or insolation (kWh/m2). It is simply a measurement of how much energy the sun provides over an area over a given period. There are several ways to measure irradiation, but one of the most important for rooftop solar panels is direct average irradiation (DNI). DNI refers to how much irradiation would be incident on a solar panel tilted so that it is precisely perpendicular to the sun (the optimal position) all of the time. It’s not a perfect representation of reality since you won’t always be perpendicular to the sun, but it gives us a standardized baseline to start with.
Your solar provider will have this data readily available for you. Still, for reference, in Illinois, the average DNI over the year will be approximately 4.8 kWh/m2/day  or 1,752 kWh/m2 per year. You can get a more accurate location in your County using the NREL Solar Radiation Database at https://nsrdb.nrel.gov.
The characteristics of your array
Once you have determined how much energy the sun is beaming down to earth at your given location, it is time to start refining that number down by accounting for specific factors to your array. The tilt and orientation of your array, along with the overall size, are the two most important variables in determining how much energy your panels will theoretically be able to harness.
The tilt and orientation of your array
We saw earlier that a homeowner could expect 1,752 kWh/m2 of solar energy (DNI) annually in Illinois. This does not necessarily equate to how much electricity the panels will produce – far from it. Recall, DNI is a theoretical value that assumes the panels are always perpendicular to the sun. Since your solar panels will be fixed to the roof, this won’t be the case, and therefore, your system will not produce the full 1,752 kWh/m2 each year.
To correct this, we must modify the DNI value to account for the tilt (i.e., roof angle) and the azimuth (north-south orientation) of the panels on your roof. This is referred to as the plane of array (POA) irradiation. A detailed breakdown of the algebra behind the POA calculation can be found at https://pvpmc.sandia.gov/modeling-steps/1-weather-design-inputs/plane-of-array-poa-irradiance/ .
Your solar provider can provide you with the tilt, azimuth, and expected POA irradiation values for your specific situation. A typical residential rooftop array in Chicago, IL, with a tilt of 20° and the azimuth of 180° (entirely south-facing) would see the 1,752 kWh/m2 DNI value reduced to approximately 1,640 kWh/m2 of POA irradiation.
The size of your array
Now that we have a much more refined estimate of the amount of sunlight hitting your panels, we must consider the system size. After all, everything we’ve been working with up until this point is on a per square meter (or square foot) basis!
Just like the amount of sunshine, it’s quite evident that the number of panels you have on your roof will directly correlate to how much energy you can produce. In the solar world, service providers will often refer to the system size, not in terms of the number of panels, but rather the panels’ total DC capacity (kWp). However, they are directly correlated to each other and to the total area covered by the panels.
A typical rooftop solar system is on the order of 5 kW and requires approximately 10 square meters per kW or 50 square meters total. Therefore, we can say that such an array would be exposed to approximately 49,000 kWh of solar irradiation over a year (1,640 kWh/m2 x 30 m2 = 82,000 kWh). The potential size of your array could be much more (or less) than 5 kW, depending on how much available space there is on your roof.
The efficiency of your system
After accounting for the tilt, orientation, and size of the array, we are now left with a reasonable estimate of how much energy will be striking the panels over the course of a year. This brings us to what many would think is the starting point: accounting for how well your system converts this energy into actual electricity!
There are two main aspects to consider when estimating overall system efficiency: the conversion efficiency of your panels and the “other” system losses after the panels start producing DC electricity.
Panel Conversion Efficiency
The conversion of the sun’s rays into electricity truly is an astonishing bit of physics. While the scientific explanation is extensive, at the end of the day the concept is relatively simple: photons (the same energy-carrying particles that plants use for photosynthesis) in the sun’s rays strike the panels and knock electrons free inside of the solar panel’s cells, creating of flow of electrons. This flow of electrons is electrical current: direct current (DC) electricity.
This process is not perfect; only 15% – 20% of the photons’ energy gets converted into electricity. So, of the 82,000 kWh striking our panels each year, only 12,350 kWh might be converted into DC electricity.
System losses account for everything outside of the panel conversion losses mentioned above. This covers dirt on the panels, shade cover, snow cover, and energy losses through the wiring.
A reasonable estimate for these losses is approximately 14% of the DC panel output .
If we take the DC panel output from the panel conversion calculation (12,350 kWh), we can see that system losses could total up to 1,729 kWh per year. This leaves a net produced amount of 10,621 kWh of DC electricity reaching your inverter.
The last step is for your inverter to convert that DC electricity into AC electricity, which, as you guessed it, introduces one more step that isn’t perfectly efficient. Typical residential solar PV inverters may have efficiencies in the neighborhood of 96%. This means that 4% of the 10,621 kWh of DC electricity is lost due to the DC-to-AC conversion inside the inverter. This leaves us with a final AC electricity production estimate of 10,196 kWh per year.
When trying to determine how much electricity can be produced by solar panels in Illinois, we must look at a few different aspects:
- How much solar energy (sunshine) there is. In Illinois, a reasonable estimate would be 1752 kWh/m2 per year.
- Angle and size of your panels. For a 5 kW system in Chicago, angled due south, we estimated 82,000 kWh per year of solar energy.
- Panel conversion efficiency. Typical panels convert sunlight into DC electricity with an efficiency of 15-20%. Therefore, our 82,000 kWh per year of sunlight translates into 12,350 kWh of DC electricity.
- Other system losses. Dirt on the panels, snow cover, shading, and inverter losses amount to another 1729 kWh of losses.
In the end, a 5 kW system in Illinois would produce approximately 10,196 kWh of AC electricity per year. This amounts to nearly the entire annual electricity consumption of the average Illinois household (10,500 kWh ). With this in mind, it’s easy to see why the residential solar market in Illinois is growing every year.