SolarSizer

What size solar system for 20 kWh per day?

Quick answer: NREL PVWatts v8 shows 1 kW of south-facing panels tilted at latitude making 1,787 kWh a year in Phoenix, 1,308 in Chicago and 1,071 in Seattle. Using 20 kWh a day is 7,300 kWh a year, so plan on about 4.1 kW of panels in Phoenix, 5.6 kW in Chicago and 6.8 kW in Seattle.

Pick your city

Array size for 20 kWh a day by city

CityPeak sun hours (kWh/m²/day)Annual AC kWh per 1 kWkW of panels for 7,300 kWh/yr (derived)
Phoenix, AZ6.571,7874.1
Las Vegas, NV6.471,7914.1
Los Angeles, CA6.091,6994.3
Denver, CO5.741,6354.5
Albuquerque, NM6.481,8174.0
Dallas, TX5.411,5134.8
Houston, TX5.151,4225.1
Miami, FL5.571,5524.7
Atlanta, GA5.091,4125.2
Nashville, TN4.771,3425.4
Kansas City, MO5.051,4515.0
Chicago, IL4.461,3085.6
Minneapolis, MN4.661,3585.4
Milwaukee, WI4.551,3375.5
Detroit, MI4.421,2875.7
New York, NY4.451,2935.6
Boston, MA4.711,3295.5
Seattle, WA3.831,0716.8
Portland, OR3.981,1136.6
Anchorage, AK3.038868.2

PVWatts columns: NREL PVWatts v8, fixed open rack, tilt = latitude, facing south, 14% system losses, standard module (full rows and request links on the peak sun hours data page). Last column = 7,300 ÷ annual kWh per kW, rounded.

How the math works

20 kWh a day × 365 days = 7,300 kWh a year. Divide that by how much 1 kW of panels makes in your city each year and you get the kW of panels. Divide kW × 1,000 by your panel's rated watts for a panel count, rounded up. PVWatts already includes its 14% loss assumption; shading, a different tilt or a different direction changes the result, so run your own address in PVWatts before you buy.

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