How to size a solar system
Divide your daily energy use by the number of peak sun hours at your location, then adjust for real-world losses from wiring, inverter, heat and dirt. The result is the DC system size in kilowatts. Divide by the panel wattage to get the panel count.
Example
900 kWh/month is 29.6 kWh/day. With 4.5 sun hours and 14% losses: 29.6 ÷ (4.5 × 0.86) = 7.64 kW. With 400 W panels that is 19.1, so 20 panels (8.0 kW).
Peak sun hours by region
| Region | Typical peak sun hours |
|---|---|
| Arizona, Nevada, Southern California | 6.0–7.0 |
| Texas, Florida, Southeast | 4.8–5.5 |
| Colorado, Utah, Mountain West | 5.0–6.0 |
| Midwest | 4.0–4.7 |
| Northeast, Great Lakes | 3.5–4.2 |
| Pacific Northwest | 3.0–4.0 |
A peak sun hour is one hour of sunlight at 1,000 W per square meter. It is an annual average; winter is lower and summer higher.
Frequently asked questions
How many solar panels do I need for 1,000 kWh a month?
With 4.5 sun hours and 400 W panels, about 22 panels (an 8.5 kW system). In sunnier states like Arizona, about 15 panels.
How much roof space do solar panels need?
A 400 W residential panel is about 21 square feet. A 20-panel system needs roughly 420 sq ft of usable, unshaded roof.
What are system losses?
Energy lost to inverter conversion, wiring, heat, soiling, shading and panel mismatch. NREL's PVWatts uses 14% as a default.