I want to install a number of Mission Solar 420 W PV panels (pdf) on my roof. By estimating an average of 5 solar hours per day over the course of the year, I believe I need 9 kW of panels. So by a nominal calculation of 420 W x 22 panels = 9240 W, or 9.24kW

However, that is only using the nominal wattage. How do I reduce that number based on roof angles and inverter efficiency to determine how many panels I actually need?

If my inverter is 95% efficient, then my usable power with 22 panels is reduced to 8778 W, so I need to add another panel to get back above 9000 W.

As for roof angles, do the average solar hours that people use for a quick calculation (5 hours) include a basic estimate of the inefficiency of standard roof angles and assume roof location can get decent coverage? Or do those calculations need to be done separately?

  • \$\begingroup\$ That 5 hour rule of thumb is meant to account for a lot of variables. It is a reasonable estimate in mid latitudes. Your actual sunlight is more than 5 hours, but the output is also less than 420 W, and the errors all cancel. Getting a more accurate estimate would probably require a lot of detailed information about your location. Best thing would be to look for output records from someone near you to see their average output vs nominal array size. \$\endgroup\$
    – user57037
    Jan 14, 2022 at 3:08
  • 1
    \$\begingroup\$ There are websites that do this for you. \$\endgroup\$
    – Andy aka
    Jan 14, 2022 at 9:18
  • \$\begingroup\$ I’m voting to close this question because this should be moved to DIY.SE \$\endgroup\$
    – Ariser
    Jan 30, 2022 at 10:48
  • \$\begingroup\$ Where are you located? - this is crucial. An excellent site to determine average kWh/m^2/day aaerage on a by monthly basis is gaisma.com. || 5 solar hours per day average across a year is very unlikely unless you live in somewhere like Kabul. Avg full sun equivalent hours tends to be rather less in winter \$\endgroup\$
    – Russell McMahon
    Feb 3, 2022 at 11:44
  • \$\begingroup\$ schinaseatribe sounds like perhaps Vietnam - yes? \$\endgroup\$
    – Russell McMahon
    Feb 3, 2022 at 11:45

1 Answer 1



PVWatts provides exactly what you're looking for.

It's a solar calculator maintained by the National Renewable Energy Lab of the U.S. Department of Energy. To use the tool:

  1. Enter the location of the proposed solar system. PVWatts will show you nearby weather stations for the simulation, and uses the most relevant solar irradiation data.
  2. Enter the system parameters. You can enter either basic info (size, orientation, angle) or provide more detail (inverter efficiency, system losses, etc). You can even draw an outline of your system on the satellite imagery of the proposed location.

The results include monthly solar radiation and AC power production values. If you included your utility rate, it will provide the value of this power as well. You can also download hourly data which includes much more detailed information on irradiance and factors such as temperature and wind speed which affect the output.

How to input your specifics

Roof angles. Entered as tilt. If there are different facets of your roof with different angles you could model each as a separate system and add it up at the end.

Inverter efficiency. Under the "Advanced Parameters" menu you can specify the efficiency of your proposed inverter if you know it, or stick with the default of 96%.

Note that there are multiple other sources of loss in a PV system as well -- PVWatts assumes total system losses of 14.08% based on the latest research. This covers multiple factors, each of which you can tweak:

PVWatts loss calculator screenshot

The help menus accessible in the interface provide a lot of good info, or for even more detail and the research that backs up the tool, you can look at the technical manual (pdf).

  • \$\begingroup\$ Thank you that was very helpful. It seems that the best way to do this would be to run a separate analysis for each section of the roof and then sum them together, since they all have different pitches and azimuths. \$\endgroup\$ Jan 18, 2022 at 19:56

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