To get kWH/day from peak kW in PV, you multiply by the average full power equivalent hours per day. In FL, this is 4 hours (mostly due to clouds). In NM the number is 5. In the continental US as a whole, the number is probably about 3.5-4. This is for a fixed (not tracking) array. This number is available on the web (I don't remember where) for anywhere in the US.
I have a 5.3 kW peak fixed PV system that provides most of the power for my house. For 6 months of the year, my electric consumption from the grid is 0 kWH or less (sometimes I have a net outflow to the grid which gets banked). FL has net metering and my system is grid-tie with no batteries. It works great. Best S. FL months are April or May. On Thu, Dec 12, 2013 at 9:43 AM, Jed Rothwell <jedrothw...@gmail.com> wrote: > Here is a graph of U.S. PV solar installations per quarter since 2010. It > shows rapid growth: > > > http://www.renewableenergyworld.com/rea/news/article/2013/12/more-records-for-quarterly-us-solar-installations > > It shows "930 MW in the July-September" quarter. That means 930 MW peak > output from the solar cells, not 930 MW of 24-hour baseline capacity. 930 > MW baseline would be the output from an average U.S. nuclear plant. I do > not know the capacity factor for solar. For wind it is roughly 30% of > nameplate capacity. > > The peak of PV solar output matches peak demand in many places, unlike > wind which tends to peak at night. > > Here is a recent graph of wind turbine output versus total power > consumption in Denmark: > > > http://www.renewableenergyworld.com/rea/blog/post/2013/12/postcard-from-the-future-122-wind-power-in-denmark > > You can see that wind is quite intermittent even on the scale of the > entire landmass of Denmark. The good news is, with today's weather > forecasting you can predict approximately how much power turbines over a > large area will produce for the next few days, so you can schedule other > dispatchable energy sources. > > - Jed > >
