One of this book's authors, Andy Friedland, has a 4,000 watt photovoltaic solar array along the side of the driveway to his house in Vermont. It consists of 16 250-watt solar panels. In 2017, this array generated 5,300 kWh of electricity. Electricity in Vermont cost $0.15/kWh. What is the capacity factor of this system and how much did Professor Friedland offset in electricity cost? If 4,000 watts were generated continuously, it would generate this much electricity in a year:
4,000 watts x q kW/1,000 watts = 4 kW
4 kW x 24 hours per day x 365 days per year = 35,000 kWh per year (after rounding to two significant figures) if the solar panels were generating 24 hours per day.

In actually, if the system generated 5,300 kWh of electricity, the capacity factor is:

5,3000 kWh/year ÷ 35,000 kWh/year = 0.15 x 100% = 15%
The capacity factor or percentage of time that the photovoltaic cells were generating electricity at full capacity was 15 percent.

To determine the dollar amount of the offset with this system, multiple the amount generated by the cost of electricity.

5,3000 kWh x $0.15/kWh = $795, Professor Friedland offset, or avoiding paying $795 to the electrical utility in 2017.

**PROBLEM TO SOLVE**:
A 3 MW wind turbine was installed in an on-land location where the capacity factor was measured to be 22 percent.
1. How much electricity will this wind turbine generate in a year?

2. How much revenue would a utility receive if they were paid $0.05/kWh for electricity?