# Cost and Area of Replacing San Onofre Nuclear Energy by Solar Photovoltaics

Nuclear power is directly emission free clean power, and replacing the fuel only uses about 5% of the energy generated, which might end up as fossil fuel usage.  In Southern California some want to replace a possibly closed nuclear plant with clean and smog free solar power. Since solar power is absent at night, when we have to burn fossil fuels, we want to add enough solar power during the day to save enough fossil fuel emission during the day to make up for that which we will burn at night.

We start with the cost part, which is a shorter calculation.  The power output of the two reactors of San Onofre combined is 2.2 billion Watts, operating steadily throughout day and night.  The average cost of building solar photovoltaic power facilities in California was about \$4.00 per Watt of peak power in mid 2011.  Since the 24 hour year round average power of these facilities is only 1/5 of peak power, to get a given average energy replacement from them requires 5 times as much peak power as you are replacing, for an energy cost of \$20.00 per averaged Watt.  Multiplying this by the steady power of San Onofre of 2.2 billion Watts, gives \$44 billion dollars, for the equivalent amount of clean energy.  By the way, the newly approved modern reactors were projected to cost \$7 billion each, so even a complete replacement of both reactors at \$14 billion would be 32% of the cost of replacing them with solar power.

For the area of photovoltaics needed, it depends on location, and whether the cells are angled at the angle of latitude for maximum efficiency.  But then you have to space them far apart to be effective near sunset when there is considerable shadowing.  So the number we get has to be taken as just a rough estimate, depending on deployment desired.  The solar insolation near Irvine from the somewhat imprecise insolation map for a tilted cell is 6kWh / m^2 / day.  This includes a 24 hour yearly average.  But now we have to multiply by the efficiency of a typical solar cell, which is only about 15%.  Since it is only a rough calculation we take the rounded result as 1kWh / m^2 / day.  In one day, San Onofre generates 2.2 million kiloWatts x 24 hours = 52.8 million kWh / day.  Dividing by the solar cell result requires 52.8 million meter squared of photo cells.  Since 1.61 kilometers equals a mile, dividing this by (1,610 m)^2 gives 20 square miles of photocells.  How big is 20 square miles?  It is ten times the size of the Orange County Great Park.

## About Dennis SILVERMAN

I am a retired Professor of Physics and Astronomy at U C Irvine. For a decade I have been active in learning about energy and the environment, and in lecturing and attending classes at the Osher Lifelong Learning Institute (OLLI) at UC Irvine.
This entry was posted in Nuclear Energy, Solar Energy and tagged . Bookmark the permalink.