Tuesday, March 17, 2009

Converting Energy to Peak Watts.

I put this out on the LDK board today. I figured I'd keep it here for posterity.

The debate starts with a claim that a company's product can put out 500MWh / acre / year, and that this is a good thing.
Well, it may be a good thing, but I can't really compare it to anything without converting it to Peak Power. So, that's what I do.

500MWh is energy, not power. So, we need to convert to Peak Power in order to compare to other systems.

Energy = Power * Time, so Power = Energy / Time.

Average Power per Acre = 500,000kWh/Year/Acre / Time (1 Year) = 500,000kWh*1day/24h*1year/365days*1/acres*1/year.

Do some cancelling and division:

The Average Power required to produce 500,000kWh in a year per acre is 57kW/acre.

Ok, so the company didn't give any idea of what assumed insolation ratio they are using here, but if it were set up in, say Arizona, and was on a dual axis tracker, 33% insolation would be a reasonable guess.

Start with Peak Power * Insolation Ratio = Actual Average Power.

Solve for Peak Power = Actual Average Power / Insolation Ratio = 57kW / .33 = 173kWp

This is the Peak Power Rating of their 500MWh/acre/year system assuming dual axis tracking, and 33% Insolation Ratio.

Comparing to a real world scenario (see).

Per Sunpower Tracker Advertising, their system works out to 161kWp/acre, which is just slightly less peak power than this reflecting system, which makes sense if the reflecting system gets a 28% conversion efficiency.

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