Maximizing the Harvest
Installing solar panels is a massive financial investment. When salespeople pitch a solar system, they often quote the "Nameplate Capacity"—for example, a 5,000 Watt (5 kW) system.
Many homeowners mistakenly believe this means the system will produce 5,000 Watts continuously, all day long. This is mathematically impossible. A 5 kW system only produces exactly 5,000 Watts in a laboratory environment, with a brand-new panel, operating at exactly 77°F, with the sun hitting it at a perfect 90-degree angle.
In the real world, atmospheric interference, heat degradation, wiring resistance, and inverter inefficiencies drastically reduce the total amount of usable electricity (Kilowatt-hours) that actually makes it into your battery bank or the utility grid.
Accurately calculating the real-world daily output of a solar array is critical to ensuring your system will actually cover your monthly electric bill.
The Importance of Peak Sun Hours
The most critical variable in solar math is geography. You cannot simply multiply your array wattage by the 12 hours the sun is in the sky.
Early morning sun and late evening sun hit the atmosphere at a severe angle, heavily filtering the light. Solar engineers use a metric called Peak Sun Hours. One Peak Sun Hour equals one hour in which the intensity of solar irradiance reaches exactly 1,000 Watts per square meter.
While the sun might be visible for 14 hours in the summer, those 14 hours might only compress into the equivalent of 4.5 Peak Sun Hours of usable energy generation.
- Seattle, Washington: ~3.5 Peak Sun Hours/day
- Denver, Colorado: ~5.5 Peak Sun Hours/day
- Phoenix, Arizona: ~6.5 Peak Sun Hours/day
How to Calculate Daily Solar Output
To find out how many Kilowatt-hours (kWh) of usable energy your roof will generate per day, you must multiply the raw size of your array by your geographic sun hours, and then subtract the inevitable system losses.
The Formula
- Determine the total Array Wattage (e.g., 10 panels × 400 Watts each = 4,000 Watts).
- Look up the average daily Peak Sun Hours for your specific city using the National Renewable Energy Laboratory (NREL) database.
- Multiply the Array Wattage by the Peak Sun Hours to find the Gross Watt-hours.
- Apply the System Efficiency Factor. (Standard grid-tied systems suffer roughly 20% in system losses due to heat and inverter conversions, leaving an efficiency rate of roughly 80%).
- Divide the final number by 1,000 to convert Watt-hours into Kilowatt-hours (kWh), which is the metric used on your electric bill.
Daily Output (kWh) = (Array Wattage × Peak Sun Hours × System Efficiency %) ÷ 1000
Example Calculation
You install a 5,000 Watt (5 kW) solar array on your roof in Denver, Colorado (which averages 5.5 Peak Sun Hours). The system operates at a standard 80% efficiency.
- Calculate Gross Production:
5,000 Watts × 5.5 Hours = 27,500 Watt-hours - Apply 80% Efficiency Loss:
27,500 × 0.80 = 22,000 Watt-hours - Convert to Kilowatt-hours:
22,000 ÷ 1,000 = 22 kWh
Your system will generate roughly 22 Kilowatt-hours of usable electricity per day. If your house consumes 30 kWh per day, you will still have an electric bill!