Wednesday, December 31, 2014

Does the math for solar photovoltaics work?

I was in a discussion about solar electricity this Christmas, and it was pointed out to me that using a solar energy of 1300 watts per square meter, and reduce that to 20% of that amount because of the efficiency of the solar cells in converting sunlight.  And the sun doesn't always shine.  The amount of solar energy is too small to be usable.

I wasn't able to answer at that time.  The math is ok, but I think the conclusion is wrong.

Here in Syracuse, NY, I averaged 17 kwh usage per day in my electric bill for this past May.  I use May because the demand for either heat or air conditioning from our electric geoexchange heat pump would be minimal.  Our loads would be lights, refrigerator, electric drier, vacuum, ironing, microwave, toaster oven, etc.  Our stove and water heater are gas fired.

I looked up the solar insolation for  Syracuse.  In January, the lowest number of the year, we average 2.8 kwh per day per meter squared, with a collector tilted at the latitude angle, 43 deg. For Syracuse.  Multiply that by 15%, the efficiency of commercially available collectors, and we have about 420 watts per day per square meter.

Following this line of reasoning, I would need 17.2 kWh/0.42 kWh per square meter = 41square meters, or 440 sq.ft.  This would be a 20x22ft. Area.  This would provide 100% of the electric demand in January, not including the building heat load.

Some critiques:
- the sun does not shine at night, so some means of storage or net metering on a grid is required.
- the months of June, July and. August have twice as much insolation per square meter, or 5.2, 5.4, 5.6, 5.3 kWh per square meter per day.
-at our house we are always trying to improve our energy performance.  We have changed our light bulbs to either fluorescent or LED bulbs.  The dishwasher and clothes washer are energy star rated.  The clothes dryer is an opportunity for improvement.  It is more than 10 years old.

I hope to get into the economics of the panel and photovoltaic electric system in a future blog.

1 comment:

  1. Please note that as the panel efficiencies increase to 20 and 25 %, the area required will shrink. 2.8 kWh per square meter x 25% = .7 kWh. 17.2/.7 = 24.5 sq.meters, or 260 sq.ft.

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