PV Fast Facts
It’s relatively easy to estimate the amount of annual output from a solar photovoltaic (PV) system, its comparative price with other solar systems, and its economics in terms of cents per kilowatt-hour (c/kWh).
Local sunlight is an important factor in the output and economics of a solar PV system. Electric output is essentially proportional to the amount of local sunlight, and cost is inversely proportional to output.
Solar maps are a good source of local sunlight. Here is one from NREL for the US, showing sunlight available for a fixed array pointing south at a tilt equal to its latitude, about optimal for such designs. You need a different map for one- and two-axis trackers for flat plates; and another one for concentrating trackers, which use only a portion of the light. But most installations are simply fixed tilt; or, in a few cases, large systems use single-axis trackers.
The NREL map shows kWh/m2-day. This is the average daily amount of sunlight on one-square meter of PV module surface during the year in these locations. To get annual exposure, multiply by 365. For example, Kansas City is about an average US solar location, with about 4.8 kWh/m2-day, which is 1750 kWh/m2-yr sunlight.
As a rough rule of thumb, a single-axis tracker will get about 25% more sunlight than a fixed array, so you could expect a single-axis tracker in Kansas City to get about 2190 kWh/m2-yr. Dual axis tracking might increase this another 8% or so. And tracking also flattens the output so it is higher most of the day, which is attractive for meeting afternoon demand.
Energy (kWh) is power over time (power – kW – times hours, if power is constant.) Instantaneous energy is power, which for sunlight varies between zero and about 1000 W per square meter at midday. If the sun shined at 1000 W/m2 for 24 hours, local sunlight could be as much as 24 kWh/m2-day! Instead, it is closer to one quarter of that throughout the US. The maximum it could be is 12 kWh/m2-day if the sun shined like it was noontime, all day every day. That’s impossible, too.
The Annual Electricity Output of a Solar System
It’s valuable to have a rough idea of what a solar array will produce. What if you install 2 kW in New York or Los Angeles? How much electricity will you get in a year?
Unfortunately, unless you design your array to be optimal, i.e., a fixed array tilted at latitude (for a nontracking array), you will have to use fairly challenging software to get a precise approximation of your output. However, you may get within 20% by the following method, which will also give you an approximate upper limit.
The approximate annual output per watt installed with no losses can be found by taking the annual sunlight in kWh/m2-yr and dividing it by 1000 W (which is the amount of sunlight on a square meter at noon). Thus annual output per watt, with no losses, in an average US location, e.g., Kansas City, is its local sunlight, 1750 kWh/yr divided by 1000 W, or 1.75 kWh/W-yr. Used in common with the US map (above) you can quickly get these numbers for most places in the US. For New York it is about 1.6 kWh/W-yr; and for Los Angeles, about 2.2 kWh/W-yr. These numbers say that for each watt of your installed array (DC rating, which is the sum of all the modules under standard conditions), if you had no losses, you’d get this many kWh of output in a year.
Unfortunately, real systems have at least about 20% losses at the busbar connection to the grid versus the sum of the module DC outputs, so the above values must be reduced by 20% from the “loss free” levels calculated above. This final amount per watt is in AC electricity. So the table shows some typical results:
Estimated annual kWh/W-installed produced in different locations (fixed flat plate at latitude tilt)
|Location||Loss Free||With 20% Losses|
|US Average (“Kansas City, MO)||1.75 kWh/W-yr||1.4 kWh/W-yr|
So if you put in a 2 kW system in one of these cities, just multiply 2000 W by these numbers to get your annual output in your location. If not in these cities, check the map, multiply by 365 to get annual kWh/m2, and divide by 1000 to get the annual kWh/W. This should get you to a rough ceiling of what you can expect. For single-axis tracking, multiply by 1.25.
The Solar Advisory Model (SAM) is a sophisticated modeling software for estimating real system characteristics, developed by NREL. But SAM can take many hours, even days to learn, and a few tiny mistakes can destroy the results, e.g., unknowingly using the wrong module technology or not changing the array tilt to match the location.
Another useful term is the capacity factor (CF). Now that you have a rudimentary estimate, you can calculate it as follows:
Annual output = CF x 8760 x Rated Output of System (Watts DC)
So if we get (for example) 1.5 kWh per year from a watt of installed PV in Miami, the capacity factor in Miami will be:
CF = Annual output / 8760 x Rated Output = 1.5 kWh/W / (8760 x 1 W) = 1.5 kWh/8.76 kWh = 17% .
Capacity factors may seem redundant, but they pop up when PV is compared to conventional energy sources, which may have CFs from under 10% for gas turbines used for a few peak demand hours a year, to baseload nuclear, used almost constantly (CF about 90%). Once you have a CF for a location, other similar systems (same module type, same array tracking or non-tracking) will have the same CF there.
Another way to quickly calculate CF is to take the local available sunlight (tracker or not) and divide it by 8760 hours. So for Miami, this would be 1900 kWh/8760 kWh = 21.7%. This would be correct except for the 20% or so losses in real systems, so the right CF in Miami will be closer to 0.8 times 21.7% = 17% (as we calculated above). Using this method to calculate CF, then multiplying by 8760 hours to give annual output per DC watt can be quite easy. Then multiply by the system rated power (DC watts) to get annual total system output in kWh. You can adjust the expected system losses up or down from 20% for your circumstances. Small, one-of-a-kind systems tend to have worse than 20% losses; some well designed big ones, less.
By the way, if you happen to have the module or array efficiency instead of total wattage, you can use maps of kWh/day-m2 (or per year) to estimate the array output by simply multiplying the daily or annual sunlight per unit area by that percentage. That then gives you the electricity per the same unit area. Mutiply by the total module area in the array to get the total output, without losses. Multiply by 80% to get an approximation with losses. So, for example, a 10% module one square meter in size might produce 10% of average US sunlight in a year, or 175 kWh/m2-yr. Reduced by losses, this becomes 140 kWh/m2-yr electricity. If the same system consumed 200 kWh/m2 to be made (including all the energy embedded in the glass and the steel and the rest), then its energy payback period would be 200/140 years, or 1.4 years (about typical for thin films).
The Economics of Solar
Dollars per Watt
One of the favorite and perhaps most annoying ways people rate PV systems is in dollars per watt. It is a very useful method if one can grasp it. The calculation itself is simple – divide the amount of money by the rated output power of the PV. This works for modules and systems, and it works for DC power and AC power, as long as properly labeled. So typical numbers one sees are: a PV module price at $2 per watt ($2/W) if a, for example, a module put out 150 W and was priced at $300; a PV module cost at $1/W (if the same module only cost $150 to make); an installed system price at $3/W (if the rest of the system price was $150, for a total of $450).
One cool way to use $/W is to estimate the impact of a component on a total system. So if an inverter costs $300,000 and it can be used to process up to one million watts (1 MW) of electricity, then its dollar-per-watt cost is $300,000/1,000,000 W = $0.3/W – about a tenth of a total system, if the system is $3/W. Another example is labor – suppose it takes 2 people at $50,000 dollars/yr per MW/yr of module output from a PV manufacturing factory. Then this is $100,000 / 1,000,000 W, or $0.1/W from this labor cost. This is about 3% of the system total if it is $3/W. 10 times more labor would be 30%, suggesting a limit to how much labor can be used to make PV. Another example: suppose it takes $30 worth of silicon to make a module that produces 150 W. Then this is $30/150 W = $0.2/W, which can again be compared to various other component costs.
Some of the detail is lost, however. For example, a tracking system will have a higher cost per installed watt than a nontracking one (due to the cost of the tracker), yet may be just as economical in terms of energy output in kWh over the course of the year, since it will produce about 25% more kWh.
We need to know what the cost per kWh of a PV system is, because this is what we pay on our bill. This is how different sources of electricity are compared, not in dollars per watt.
You cannot do this precisely without using software like SAM, because cost per kWh is highly dependent on several subtle variables: not only cost per watt and annual output, but financial assumptions like loan rate and duration; and taxes, including federal and state income and property tax. There is also a small contribution from operating and maintenance costs, even though some of these systems are run without anyone on-site. Insurance costs and connecting to the local grid also contribute their share.
So we cannot say anything universal, but we can make usable estimates.
To first order, in the US Southwest (defined as a location where there is 2400 kWh/m2-yr, or a CF of 22%), each dollar per watt of PV system cost is equivalent to about 5 c/kWh of cost. This is what you get if you run the software with reasonable assumptions.
So if a system is $2.5/W installed (counting all costs), it will produce electricity at about 5 x 2.5 = 12.5 c/kWh. (Remember, each dollar per watt is equivalent to about 5 c/kWh, so this is 2.5 times that.) Then if you want to adjust this amount to other locations in the US, take their local sunlight and create a ratio, sunlight in the SW/sunlight at new location, and multiply this by 5 c/kWh per dollar a watt to get a new amount of cost per dollar a watt in that new location.
An example helps. Say we wanted to put our array in New York City, with 1600 kWh/m2 yr. Then the ratio is 2400/1600 = 1.5. This is how much more PV electricity costs in NYC. So if the system were 12.5 c/kWh in the SW, it would 1.5 x 12.5 c/kWh = $0.19/kWh in NYC. (Funny thing is, this is about what people pay for retail electricity in NYC, so that’s something to consider.)
If you want, you can do this then add another 1 c/kWh for ongoing O&M to get a slightly better estimate.
If you have a tracking array, you can use the same adjustment: just use the same ratio of SW sunlight to new location sunlight, with the tracking sunlight in the denominator. Since it may very well be higher than 2400, it will reduce the cost of the system. So the ratio of using a tracker to not using one in the same location would be X/X*1.25 = 0.8, if the capital cost stayed the same (which it won’t). Or you can think of each dollar-per-watt of tracking array cost as implying only 4 c/kWh costs (20% less, since X/1.25X = 0.8).
Given these simplified formulas, you should be able to estimate out how much PV costs anywhere in the world, as long as you know the local sunlight on a fixed flat plate, tilted at latitude. To help that along, here is the world solar map. This one is in average power over 24 hours (including at night, kind of an odd idea). To get local daily kWh as in the figure at the top, multiply by 24; so 300 W/m2 becomes 7.2 kWh/m2-day. Then multiply by 365 to get kWh/m2-yr.
According to the source, the black dots taken together would supply all the world’s energy.