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Subject:

Re: Help! I Have Solar Power Questions

From:

Nathan Mitten <[log in to unmask]>

Reply-To:

Nathan Mitten <[log in to unmask]>

Date:

Thu, 13 Mar 2008 13:21:13 -0400

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 text/plain (110 lines) , image002.jpg (110 lines) , Regional per capita solar electric footprint for the US.pdf (110 lines)
 ```Hey Patrick, I hope you have some answers to provide to the planning council staff. I'm not sure how much we helped you. As Dr. Humphrey stated, there are many different efficiencies associated with the different current technologies and they are constantly changing. Here is a step by step method to calculate system efficiency and land use for a PV system with more detail:   1) find the PV module efficiency which is normally measured as DC Watts output per meter squared / 1000 Watts per meter solar input (i.e. - 15%)   2) calculate the derate factor to go from panel efficiency to system efficiency which is the product of all other system component efficiencies (i.e. - 77% see http://rredc.nrel.gov/solar/codes_algs/PVWATTS/system.html for details)   3) The product of these gives the system AC efficiency (i.e. - 11.6%)   4) To get AC power rating multiple this by 1000 W/m2 (i.e. - 116 WattsAC/m2)   5) This is the power rating for a flat mounted system, to get land use this needs to be multiplied by an effective land use factor, these are based on NREL's report on PV land use from Dec of 2007 (table below) which I'll attach, see note at bottom of page. Multiply the AC power rating by: Flat (roof): 1 10 deg, south facing (roof): 0.87 (i.e. - 101 WattsAC/m2) 25 deg, south facing (ground): 0.48 0 deg, 1 axis tracking: 0.35 2 axis tracking: 0.15   6) Now divide the total AC system size (i.e. 200kW) by the AC power rating (i.e. 0.101 kWattsAC/m2 for 10deg,south) to get total meters squared for the installation (i.e. 1980 m2)   7) Now divide by 4046 m2 in an acre to get acres needed for that AC system size (i.e. 0.49 acres for a 200 kw AC system or about 400kW AC / acre)   8) To get actual energy production per day, multiple by ~5 sun hours/day for Florida, see page 24 in the attached document for other state or various mounting configs (i.e. 2000 kWh / day / acre or 0.73 MWh / year / acre)   Now this process should work for any panel efficiency, system derate factor, capacity factor (or sun hours/day), and any mounting configuration. Hope this helps, Patrick.   Note: As can be seen below in the table from the NREL report, tilt and tracking can dramatically improve your total energy output but it drastically increases the amount of land you need do to shading.       Table 1 illustrates the significant drop in PV array power density for tilted and tracking arrays due to shading and maintenance requirements. This drop in power density is accompanied by a greater energy yield per installed unit of module power. However, the reduced power density is much greater than the increased collector yield, so moving from flat rooftop arrays to land-based tilted and tracking arrays can reduce system energy density by more than 50%. Improvements in system energy density will be driven more by module efficiency increases than by improved array spacing because shading and maintenance requirements provide fundamental limits on array packing density, while deploying more efficient cells can substantially improve system energy densities in the future.24      BEST, Nate Mitten   ```

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