Sunday, January 11, 2009

A Note on Units of Energy and Insolation.

This post is in reference to my use of units like "Watt*1Year," or "Watt*25Years," etc, in posts such as This, This, and This, and might just be useful in laying out some of the basic math behind Solar Energy Output. Feel free to critique.

In Physics, Power is described in Watts. Energy is described by Power * Time. Typically when we think Electrical Energy, we think in terms of Kilowatt*Hours, but the actual units used for Time are arbitrary, it's just a matter of the increment of time over which you are considering the flow of Power.

The Quantity of Energy streaming down on the planet can be measured in terms of its Insolation. The problem that I'm seeing out there is that it's not firmly decided what units we should be using for Solar Insolation, and there is little way to translate at a glance quantities from one choice of Unit to the next. Now, maybe there's a reason that somebody would want to use "kW·h/(m²·day)" or "kWh/kWp•y" for practical applications to Solar Energy, but the rationale certainly escapes me. What I do know is that a Solar Module is rated in Watts Peak (Wp), which is the Power Generated when the Panel is exposed to an Insolation of 1000W/m2. So, to match this, I want my units to be comparable to W/m2.

Following is a map of US Annual Insolation in kW/m2*.

By taking the given values in terms of kWh/m2/day, converting from KiloWatts to Watts, and multiplying each by 1day/24h to cancel out the elements of time, we get the Annual Average Power, in W/m2. Once we know this Annual Average Power, then by dividing it by the 1000W/m2 rated Peak Power used by the Photovoltaic Industry, we get a very useful percentage.

Example: Looking at the map, let's take a spot on one of the bright yellow areas, like is found in most of Virginia. The legend shows an Insolation Value of 4.5-5kWh/m2/day. Converting to Watts, and taking the range's lowest value of 4500Wh/m2/day, multiplying by 1day/24h, and canceling out the hours and days, gives 187.5W/m2.

So, now that we have the average Rated Insolation for the location, then we divide this number by 1000W/m2 in order to get the Actual Insolation as a percentage of Rated Peak Insolation, in this case, for Virginia, at 18.75%.

I've run this calculation for the various brackets in the map legend, and have added these percentages to the graphic. The spreadsheet is here.

Lets say that you want to know roughly how much actual Energy some Solar Installation will produce over a year. You just take the Peak Power rating of the Installation, and multiply by the Percentage that was calculated above, and then multiply by 1Year in order to get the Energy produced on average over that Year.

Example: You want to know how much Energy is going to be produced over the year by a 5kWp Installation in Virginia where the expected average Insolation is 18.75% as calculated, above. Simply take the Peak Rated Power of the Installation, and multiply by the Percentage and 1Year, in this case, 5kW*18.75%*1Year = 937W*1Year.

This is the Average Energy Produced over a Year for this Installation, even though it's not in the usual units. To convert to kWh, just convert the Year to Hours using the factor of 8760Hours/Year and 1kW/1000W to get 8208kWh.

Now let's say that you want to make a comparison in Cost per Watt between a Solar Installation and a Coal Plant, or a Natural Gas Turbine, or any other conventional Electrical Generator running at a Constant Output over the year. Just remember that the total Energy Produced by a constant generator over a year, in W*1Year (or kW*1Year, or GW*1Year), is roughly it's Rated Output * 1Year, so a 100MW Coal Plant should produce in the area of 100MW*1Year in Energy over the year. We could convert this to kWh just like was done above for the Solar Installation, but there's no need to do so if we're just using it for comparisons-sake.

* This map measures Insolation assuming optimally angled panels, so for flat-roof installations, particularly at higher Latitudes, will over-estimate output. For a European Map and Insolation Values that assume flat placement of Panels, see Lightbucket.

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vitaliyinc said...

I do it a bit differently by figuring in PV efficiency.

So, in Virginia we have 4.5kWh/m2/day average insolation onto a stationary flat-plate collector facing south tilted at lat angle.

So a 5kW solar panel setup with an efficiency of 12.5% would be about 40 m2 of area (40 125W panels).

Therefore, annual production = (4.5kWh/m2/day)*(365days)*(40m2)*(0.125) = 8212.5kWh (or 937.5W avg)

The 8212.5kWh would change if the panel efficiency is different from 12.5% or if the inverter inefficiencies are figured in.

Anonymous said...

kWh/m^2/day is a very useful measure if you're looking at off-grid systems with a constant load (or assumed constant load).

Then you simply use your daily demand (in Watt-hours), an efficiency factor, and the insolation in kWh/m^2/day to give you the required Wp rating of the array.