Solar system size calculator guide
What does this calculator help you figure out?
Solar system size should connect the electricity you use each day with the solar resource at your site and the losses between panel and meter. This calculator takes daily electricity use in kWh, the local peak sun hours, a system-loss percentage covering inverter, wiring, soiling, and temperature effects, and an optional orientation or shading loss, then returns the nameplate capacity in kW needed to produce that daily energy on an average day. If you enter an available roof area it also shows a rough capacity check using a planning density of 0.2 kW per square metre, so you can tell whether the required array is likely to fit. The result is a planning estimate for conversations with installers, not a production guarantee: seasonal variation, snow, shading, panel choice, and export limits all move real output. Use your utility bill for daily consumption and a resource map such as NREL’s for peak sun hours.
How is the result calculated?
Base size (kW) = daily energy use (kWh/day) ÷ [peak sun hours × (1 − system losses ÷ 100)]. System size (kW) = base size ÷ (1 − orientation/shading loss ÷ 100). Rough roof capacity (kW) = roof area (sq ft) × 0.092903 × 0.2.
Worked example
Worked example with the defaults: a home using 20 kWh per day in a location with 4.5 peak sun hours and 20% system losses. The loss factor is 1 − 0.20 = 0.80, so each kW of panels yields 4.5 × 0.80 = 3.6 kWh per day. The required capacity is 20 ÷ 3.6 = 5.56 kW, and with the default 0% orientation loss the calculator displays 5.6 kW. The loss-adjusted output row confirms the sizing: 5.56 × 4.5 × 0.80 = 20 kWh/day. Adding a 10% shading loss would raise the size to 5.56 ÷ 0.90 = 6.2 kW, and entering a 400 sq ft roof would show a rough capacity of 400 × 0.092903 × 0.2 = 7.4 kW.
Units and conversion notes
Use kWh/day for consumption (divide a monthly bill’s kWh by the days in the cycle), peak sun hours per day for the solar resource, and percentages for both loss fields. Peak sun hours are the equivalent hours of 1,000 W/m² irradiance, typically 3–6 in the continental United States, not the hours of daylight. Roof area is in square feet and tilt in degrees; tilt is shown for the site review and is not converted into a production factor. Roof capacity is a rough planning check, not a structural approval.
What does the result mean?
The result is a nameplate planning estimate, not a promise of monthly production. Seasonal weather, shading, roof orientation, export limits, and equipment choice need a site review, and winter output can be half the annual average.
Common mistakes to avoid
Do not divide by daylight hours instead of peak sun hours; a 12-hour day at a site with 4.5 peak sun hours would undersize the array by more than half. Do not treat nominal panel power as daily energy, and do not apply losses twice by entering inverter losses in both the system-loss and orientation fields. Do not size to a summer bill alone if you want year-round coverage, and remember that the 0.2 kW/m² roof density assumes standard-efficiency modules laid edge to edge, without the setbacks, vents, and access paths a real roof requires.
Use a local production study and qualified designer for procurement, structure, wiring, and interconnection decisions.
Common questions
Frequently asked questions
How many kW of solar do I need?
Divide daily use by local peak sun hours and adjust for losses. A home using 20 kWh a day with 4.5 peak sun hours and 20% losses needs about 5.6 kW; the required size changes with season, shading, and whether you want to cover all or part of your bill.
Does roof area determine system size?
Roof area can limit the practical array, but orientation, setbacks, structure, and panel layout also matter. At a rough 0.2 kW per square metre, a 400 sq ft (37 m²) roof section supports about 7.4 kW before setbacks, so it would fit the default 5.6 kW system.
What are peak sun hours?
Peak sun hours express the day’s total solar energy as equivalent hours of full 1,000 W/m² sunshine. A site with 4.5 peak sun hours receives 4.5 kWh/m² per day on average, even though the sun is up for 10–14 hours; NREL resource maps publish the figure by location.
What system losses should I assume?
A 14–20% total loss is a common planning range, covering inverter conversion, wiring, soiling, temperature, and mismatch; NREL’s PVWatts tool defaults to 14.08%. Use 20% for a conservative first estimate and lower it only with installer data.
How many panels is a 5.6 kW system?
Divide system size by panel power. With 400 W panels, 5,600 ÷ 400 = 14 panels; with 350 W panels it is 16. Use the solar panel count calculator to match a specific module and check the roof layout.
