Solar Power Converter

Solar panel energy output depends on panel wattage and sunlight hours.

Formula:

• Daily kWh = Panel Watts x Sun Hours x Efficiency / 1000
• Yearly kWh = Daily kWh x 365
• Monthly kWh = Yearly kWh / 12, which is Daily x 30.42

That last one is worth writing out. Multiplying a daily figure by a flat 30 loses five days a year, so the monthly and yearly numbers stop agreeing with each other. A twelfth of a 365-day year is 30.42 days.

Typical efficiency factor: 0.75-0.85 (accounts for inverter losses, temperature, dirt, etc.)

Average peak sun hours by region:

• Arizona/Nevada: 6-7 hours
• California/Texas: 5-6 hours
• Midwest US: 4-5 hours
• Northeast US: 3.5-4.5 hours
• Northern Europe: 2.5-4 hours
• Australia: 4.5-6.5 hours

Common panel sizes:

• Residential panel: 300-400 watts
• Typical system (20 panels): 6-8 kW

Example:

• 400W panel x 5 sun hours x 0.80 efficiency = 1.6 kWh/day
• 20 panels = 8 kW system = 32 kWh/day = 973 kWh/month = 11,680 kWh/year

That monthly figure is 11,680 / 12, not 32 x 30. The flat-30 version gives 960, and this page used to print it two lines under the paragraph explaining why that is wrong.

The single most important idea here is the difference between power and energy, the thing people mix up constantly. Watts measure power, the rate of generation at an instant; watt-hours measure energy, the total produced over time. A panel’s wattage is its peak output under ideal lab sunlight, not what it makes across a real day.

Actual output is always lower, because the sun isn’t at full strength for more than a few hours. That’s why solar production is figured from “peak sun hours,” which compresses a day’s changing sunlight into an equivalent number of full-strength hours. A 400 W panel in a region with 4 peak sun hours makes roughly 1.3 kWh a day at 80% system efficiency, not the 9.6 kWh it would make if it ran flat out around the clock.

Sizing a system backwards

The useful direction is the other one, and the calculator above does it: start from what you actually use. Take your daily consumption in kWh off your electricity bill, divide by your local peak sun hours, divide again by the system efficiency, and you have the array wattage you need.

A household using 30 kWh a day in a 5 sun hour region needs 30 ÷ 5 ÷ 0.80 = 7.5 kW of panels, which is about 20 modern panels. Notice how hard the efficiency factor works there: at 0.70 the same house needs 8.6 kW.

One thing the yearly figure hides

A single sun-hour number averaged over the year is fine for annual planning and misleading for anything monthly. December in the northeastern US delivers perhaps a third of what June does. If you are sizing for winter self-sufficiency rather than an annual bill, use your December sun hours, not your average, and expect the array to be roughly twice as large.