Solar domestic hot water:
The average person uses approximately 20 gallons/day of hot water for domestic water needs (showers, laundry, cooking). Many families benefit from a solar domestic hot water system that includes (2) 4’ x 8’ solar thermal collectors and an 80-gallon solar storage tank. Any additional hot water needs would come from a back up heating element using electricity, natural gas, or propane.

Off-grid solar electric:

When you are sizing a solar electric system for off-grid use for places where there is no utility grid source of electricity, it is important to carefully analyze your electricity needs, because you do not have an easy grid back-up, your back up power needs to be stored in batteries. The first step is to make a list of your appliances and their electricity draw, including how many hours/day you typically use them. For an off-grid scenario you also want to consider how involved in your energy consumption you are willing to be. If we have a few weeks in January with day after day of overcast skies, are you willing to curtail your energy consumption in order to help your battery power last, or do you have electricity needs that are not flexible, and therefore a larger battery bank is required? Sometimes a second backup source (generator) is also used.

Grid-connected solar electric:
When your solar electric system is connected to the grid, there is no set rule for how big your solar electric system needs to be, because you can make 100%, 50%, or even less of your electricity from the sun, and still have back-up ‘regular’ utility power as needed.

Converting KW into kilowatt-hours:

A 1 kilowatt (KW) solar electric system includes approximately (5) 200 watt photovoltaic modules.
(5) x 200 watts = 1,000 watts or 1 kilowatt (KW)
This 1 kilowatt system would make 1 kilowatt-hour of electricity for every hour it is in the sun.
In Minnesota we have an average of 4.6 good sun hours every day (more in the summer when the days are longer, less in the winter when the days are shorter).
1 kW solar electric system x 4.6 sun hours/day = 4.6 kilowatt hours/day of electricity.

4.6 kilowatt hours/day x 30 days/month (average) = 138 kWh/month of electricity from a 1 kW system. (x 12 = 1,656 kWh/year). Remember these are averages, based on compiled weather data, and not including any losses for inefficiencies, wiring, inverter losses. Often we expect a photovoltaic system to perform at about 80% efficiency, so 1 kW solar electric system would, in reality, produce ~1300 kWh/year.)

You can use this 1 kW increment to figure out how much solar it would take to offset 100% of your electricity use.

Your kWh use (per month) ÷ 30 days = your kWh use/day
Your kWh use/day ÷ 4.6 sun hours =  your kW needs to offset 100% of your electricity.
Your kW needs ÷ .8 to account for inefficiencies = the size of your PV system.