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OWOE - Solar Power - What is the average size of a rooftop solar system?
  Figure 1 - Photovoltaic Solar Resources - Average Annual Insolation Values (NREL)
Figure 1 - Photovoltaic Solar Resources - Average Annual Insolation Values (NREL)
Figure 2 - Annual Insolation Chart for Los Angeles (Solmetric)
Figure 3 - Average size and efficiency of solar modules (Solar Power Rocks)
What is the average size of a rooftop solar system?
Topic updated: 2024-03-18

The three key input values to determine how large a rooftop solar system needs to be are:
  1. Insolation value for roof location - Insolation is the amount of solar radiation that hits the earth and is measured in energy (i.e., watts) per area. Although 1367 watts per square meter (W/m2) of sunlight strikes the outer atmosphere, about 30% of it is reflected back into space. Therefore, when the sun is directly overhead, the solar energy that passes through the atmosphere and strikes a perpendicular surface at sea level is about 1000 W/m2. This amount varies with latitude, altitude, season of the year, and weather. The map generated by the National Renewable Energy Laboratory in Figure 1 can be used to estimate insolation across the US. It shows the average annual insolation in kW-h/m2 per day for a fixed solar system that is aimed to the south and tilted toward the sun at the same angle as the lattitude of the location. One can see that for most of the southeastern and western US and Hawaii, excluding the southwestern deserts, an average insolation value of 5 kW-h/m2 per day is a reasonable assumption. This is interpreted as these locations receiving the equivalent of 5 hours per day of full sunlight over a 24 hour period (or Equivalent Hours of Full Sun (EHS)), accounting for yearly variations and typical cloud cover. To get more specific data on specific locations and panel orientation, a look-up tool such as Solmetric can be used. Figure 2 shows an insolation chart for Los Angeles using this tool. At the optimum tilt (31 deg) and azimuth (190 deg) yearly insolation is 2065 kW-h/m2. This gives an average daily insolation of 5.65 kW-h/m2.

  2. Panel size and rated power - All solar panels are rated by the amount of DC (direct current) power they produce under standard test conditions. Based on 2018 data collected by the State of California (see Figure 3), the typical solar module is 39" wide by 65" tall (17.6 sq.ft. or 1.6 sq.m.) and has a capacity rating between 285-360 watts. This is based on an average efficiency of 18.7% and using the sun's full insolation capacity of 1000 W/m2) . Using an average annual insolation value of 5, a 320 watt solar module would generate about 1.6 kW-hours on average every day during the year. This value must then be downrated by 10-20% to account for system losses and the conversion of DC power to AC power.

  3. Average dialy electricity usage - Per the Energy Information Administration (EIA), in 2021 the average home in the United States used between 6,400 kW-h of electricity per year (Hawaii) and 14,300 kW-h per year (Louisiana), with the average of approximately 10,600 kW-h per year.
Using the above values and a 15% system loss, the average home in Hawaii would require 1.15 x (6,400 / 365) / 1.6 = 13 panels = 229 sq.ft. (21 sq.m.) with a rated capacity of 4160 kW; the average US home would require 1.15 x (10,600 / 365) / 1.6 = 21 panels = 370 sq.ft. (34 sq.m.) with rated capacity of 6720 kW, and the average home in Louisiana would require 1.15 x (14,300 / 365) / 1.6 = 28 panels = 493 sq.ft. (45 sq.m.) with rated capacity of 8960 kW.

There are a number of other critical considerations. On the positive side, a more sophisticated (and costly) system that tracks the sun could result in a 30% increase in insolation and similar decrease in required panel area. Conversely, a system that cannot be aligned to the south or tilted at the same angle as the location's lattitude (i.e., due the orientation or pitch of the roof) will receive less insolation. It should also be noted that this amount of solar modules only covers the average daily power usage during the year. The system would be underpowered for days with higher needs, greater cloud coverage than average, nighttime, etc. Conversely, it would be overpowered for days with lower needs, less cloud coverage, peak sunlight, etc. A system would need to be oversized and/or depend upon grid power to make up the additional needs. And, finally, a completely off-grid installation that depends on battery storage for 24-hour power would be subject to another 30% efficiency loss.

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