Solar power efficiency? Just to make sure I understand this solar power thing:

1. Peak solar constant on the earth%26#039;s surface = 1 kw/meter
2. Photovoltaic efficiency = 20%
3. Intermittency efficiency = 20%

(this is the relation of the peak to the average unit time, totalling across day and night, good weather and bad, summer and winter)

4. Storage efficiency = 25% (Storage efficiency comes in two parts: losses inflicted by putting your power in storage and losses inflicted when taking it out. I am assuming 50% in both cases.)

So we end up with 1000 kw/m * .2 * .2 * .25 = 10 watts a sq meter.

I%26#039;m not saying that%26#039;s good or bad. I%26#039;m just asking if it is right.

Asker's Rating:

helpful.

Other Answers (2)

• The efficiency is closer to 30% on new unit 36% is a stretch. On the average in the US, 4.5 hours per day average of peak, that%26#039;s 20%.
  
  Batteries and inverters lose 10% in and 10% out, net 80%, but that%26#039;s peak, so a net 85% eff. The sun averages 1.2 kW/m^2, so .2*.3*.85*1200 = 1.47 Kw-hr/day per m^2 of panel. This is worth $.22 per day of use. So 10 m^2 is worth $800/year
• current Sharp solar cells have a 36% efficiency.
  
  I get 4.94 kWh/sq-m/day here in Louisiana
  
  
  not really sure about the Intermittency value.