Solar power efficiency?

Question

Just to make sure I understand this solar power thing:

1. Peak solar constant on the earth'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'm not saying that's good or bad. I'm just asking if it is right.

Answer 1

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's 20%.

Batteries and inverters lose 10% in and 10% out, net 80%, but that'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

Answer 2

The photovoltaic efficiency should be lower than 20% for a practical system.  The most expensive widely-distributed cells, Sunpower, have an efficiency of maybe 22%.  Put put them under glass in a practical frame, and it's down to 19%.  These are $7 a watt type panels.  The $4 a watt panels that most folks install are 13-15% efficient, after they're put in a practical frame.  The 36% type of cells from places like Spectrolab are $250 a watt - not for homeowners.

Intermittency efficiency looks good for a place like California.  What you're saying by that 20% is that you're expecting 24 x .2 = 4.8 equivalent peak-sun hours year round.  This number should be higher in Southern Cal or Arizona, lower in Michigan.  Really, this number can be the normalization factor that takes into account all the other aspects of the area - atmospheric transparency due to altitude, pollution, operating temperature.

For storage efficiency, I think you can get better than 25%.  I'd use 70% - 80% for the battery, and 95% for the inverter, if any.

Now, for the real-life number from our system (see link below):

6556 (kWh/yr) / 8765 (hours/yr) / 12.5 (square meters of panel) = 60W / m^2, approximately.  We are grid-tied, no battery losses.

Answer 3

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.