What the animation shows
The animation and the step-by-step video above use your numbers. Here is the same walkthrough written out for the default example.
What you enter
- Average monthly electricity use: 900 kWh.
- Share of use to cover with solar: 100 %.
- Peak sun hours per day (annual average): 4.5.
- System losses: 14 %. 14% is the PVWatts default (wiring, inverter, soiling, shading, etc.).
- Panel rating: 400 W.
- Area per panel: 21 sq ft.
What happens, step by step
- Your site gets 4.5 hours of full, peak sunlight a day.
- Each 400 W panel makes about 565 kWh a year after losses.
- Panels go on until they cover 100% of 10,800 kWh a year.
- That takes 20 panels, or 8 kW of solar.
The answer: system size is 8 kW.
How it’s calculated
- Target kWh/yr = monthly use × 12 × offset
- kWh per kW per year = peak sun hours × 365 × (1 − losses)
- Panels = target ÷ yield per kW ÷ panel kW, rounded up
- Production = installed kW × yield per kW
Worked example
900 kWh per month, 100% offset, 4.5 peak sun hours, 14% losses, 400 W panels.
Inputs
- Average monthly electricity use
- 900 kWh
- Share of use to cover with solar
- 100 %
- Peak sun hours per day (annual average)
- 4.5
- System losses
- 14 %
- Panel rating
- 400 W
- Area per panel
- 21 sq ft
Results
- System size
- 8 kW
- Panels
- 20 panels
- Panel area
- 420 sq ft
- Estimated yearly production
- 11,300 kWh
Assumptions and limits
- Peak sun hours are for the actual tilt and orientation of your roof.
- Rounding up to whole panels means production slightly exceeds the target.
- Estimates are not guarantees; weather varies year to year.
Verify the result against the applicable local code, adopted code edition, manufacturer instructions and project requirements. See our methodology.
Questions
How many solar panels for 900 kWh a month?
With 4.5 peak sun hours and 14% losses, about 7.6 kW — twenty 400 W panels.
Where do I get peak sun hours?
NREL’s PVWatts calculator estimates production for your exact address, tilt and azimuth.