In a previous post, I reported that French journalist and economist Guy Sorman proclaimed, “If California were to rely on solar power for its electricity consumption, the entire state would have to be covered with photovoltaic cells.” In reply, I proclaimed that Mr Sorman is wrong by 50,000%.

How would you go about deciding who is correct? In teaching and observing solar classes and talking to a variety of people outside the solar industry, I’ve found that many or even most people can’t with confidence confirm or deny this type of claim.

So let’s go through it. It’s a bit laborious, but the good news is it’s not technically difficult. Even better news is that the few key concepts needed to dig into the claim are probably also the most important needed to start thinking through solar on your own rooftop.

Kilowatts (kW) and kilowatt-hours (kWh) – these are very clearly defined units, but also very frequently confused. kW is a unit of power, kWh is a unit of energy. Power and energy are unambiguously different from physics and engineering perspectives, but are often used interchangeably in common English as well as by some reporters who should know better. Unfortunately, many web pages define watt in potentially confusing ways such as in terms of other more fundamental units of physics. This doesn’t help. Also, one of thefreedictionary.com’s 16 definitions of power is “Electrical or mechanical energy.” This *really* doesn’t help.

**A good definition of power is: the rate at which work can be done.** A good example is a 100 watt incandescent light bulb. If it’s on, the work it’s doing shows up as light and heat. It’s designed to run at 100 watts and no more, and and so its name is based on the peak capacity of the bulb to put out light and (a lot of) heat. Solar panels are also named by the maximum amount of watts they can put out. A solar panel is kind of the opposite of a light bulb: Put light into a solar panel and out comes electricity (and some heat).

Like light bulbs, solar panels come in different wattages. A common power rating for a high end solar panel is 345 watts. The size of this panel is about 61″ by 41″ or about 17.3 square feet. So, this panel, at its maximum, puts out 345 watts from sunlight falling on its 17.3 ft² area. Another way to say this is, at its maximum, **a 345 watt solar panel puts out a maximum of about ****20 watts per square foot** (345 divided by 17.3 equals about 20).

Back to the light bulb. If a 100 watt light bulb is left on for one hour, it will consume 100 watt-hours of energy. Left on for 10 hours, it will consume 1000 watt-hours, which is the same as 1 kilowatt-hour, or 1 kWh. Similarly (and under ideal conditions), if a 345 watt solar panel is left in the brightest sun for 1 hour, it will generate 345 watt-hours of energy. Under those same ideal conditions, after three hours, it will generate a little over 1 kWh.

One might wonder if “watt-hours” might mean watts minus hours (since it looks like that) or watts per hour, or some other variant. But watt-hours means watts multiplied by the hours the watts are doing work. The definition of energy is very clear to physicists, but there are probably even more English language definitions for energy than power. **A good definition of energy is: power consumed (or generated) over time.**

The size of a solar system is specified in watts, or kilowatts (kW) or megawatts (MW). For example, a commonly seen residential rooftop solar system might be 4kW, a large solar power plant in the California desert can be well over 100 MW. The output of a solar system is given in kilowatt-hours (kWh) or megawatt-hours (MWh) or gigawatt-hours (GWh). Only the W is “supposed” to be capitalized but few people care and all variations are used interchangeably. Your electricity bill is priced in cents per kWh. There are many pricing complexities; in California you’ll probably be paying somewhere between a little under 10¢/kWh to (rarely) over 50¢/kWh.

The big question, whether you’re considering solar for your roof, for all of California, or for anything in between, is how much energy will your system generate? How many kWh’s will your kW’s generate? There are very many factors that go into making the calculations, including some, like the weather, that are not precisely predictable. A lot of effort goes into the design, construction and monitoring of solar systems, to make sure that the kWh’s that are supposed to be generated, actually are. But the good news it’s easy to arrive at ballpark numbers. Estimates made by competent people off by over 10% are very unusual. (I am claiming that Mr Sorman is off by 50,000%.)

So… how do you estimate kWh’s from kW’s? Here is just one way to do it:

- Step 1 is to start with the maximum power your system can generate, and then “de-rate” it for each less than ideal factor. For example, dust may build up on the panels, blocking some of the light. Also, the “direct current” (DC) that the panels produce must be “inverted” into “alternating current” (AC) that homes and the power grid use, and there is some loss in the process. There are other factors that every solar installer understands, and there are standard tools to do the calculations. The result is that the maximum AC power that a rooftop solar system will generate is typically between 75% and 80% of the maximum DC (“nameplate”) power that’s stamped on the panels by the manufacturer. For example, if you purchased a 4.14 kW system (that would be 4.14 kW DC, comprised of 12 345-watt DC panels), it would generate at its maximum between about 3.1 kW and 3. 3 kW AC. Let’s go with 3.1 kW AC.
- Step 2 is to calculate how much energy the system will generate on an average 24-hour day. At noontime on the brightest summer day, the system will put out 3.1 kW. Over the noontime hour on that bright day, it will generate 3.1 kWh. An average day means taking the average of a full year’s worth of 365 days to account for seasonal changes. This would be a hard calculation involving morning fog, nighttime, rain, the sun’s varying angle, but the National Renewable Energy Laboratory (NREL) has made it easy for everyone, by giving us the “peak sun hours” for a large number of locations across the country. Peak sun hours describes, given your system’s maximum power (3.1 kW AC), how much energy that system will generate on the average day. The San Francisco Bay Area’s number is about 5.5, meaning for a 3.1 kW AC system, 3.1 times 5.5 or about 17 kWh will be generated during the average day. Because the 5.5 is based on the full year’s worth of 24-hour days, the yearly output of the 3.1 kW AC system is simply 365 times 17 kWh or 6205 kWh.

Now we can figure out how much energy (kWh) will be generated per square foot of solar panel. The 3.1 kW AC system is a 4.14 kW DC system made up of 12 345-watt panels, where each panel is about 17.3 ft². So 12 panels would be about 207 ft² and if that 207 square feet of panels generates 6205 kWh per year, then **one square foot of solar panel generates about 30 kWh per year** (6205 divided by 207 = 30).

Just two more things and we’ll be able to assess Mr Sorman’s claim: how big is California, and how much electricity does California consume in a year.

- Enter “size of California” into Google:
**California is 163,696 square miles**. That was easy.
- Electricity consumption is a little more involved. It’s under the California Energy Consumption Data Management System here. Choose “ALL” under County, select “Total” in Sector, select the most recent year (2011), click on Create Report. “Total” reports all types of solar per county, but does not give the sum of all the counties which is the total for California. So copy the entire report (column titles plus 58 counties) and paste it into a spreadsheet. Then add up the county totals with a formula such as “sum(c2:c59)” and the result is 272645.3171. The report says all numbers are in expressed in millions of kWh, so we now know
**California consumed 272,645 million kWh of electricity in all of 2011**. Other ways of expressing 272,645 million kWh include: 272,645 thousand MWh; 272,645,000 MWh; 272,645 GWh. We can get a sanity check in the California Energy Commission’s Energy Almanac here where it says California generates about 200,000 GWh per year and that’s about 70% of how much California uses. 200,000 GWh divided by .7 = about 285,000 GWh which is a good match to our 272,645 GWh.

How many square miles of solar panels would be needed to generate 272,645 million kWh per year? The arithmetic is simple, but it’s with giant numbers so care is needed. Our one square foot of panels generates 30 kWh per year. There are 5280 times 5280 or about 28 million square feet in one square mile. So one square mile of panels will generate about 840,000,000 kWh per year (28 million times 30 kWh) which is the same as saying **one square mile of panels will generate about 840 million kWh per year**. Now we can just divide total consumption by generation per square mile to find how many square miles we need to satisfy California’s consumption. 272,645 million kWh divided by 840 million kWh = 324. So **324 square miles of solar panels will generate enough electricity during the year to satisfy California’s total electricity consumption for the year**.

Last step! California has 163,696 square miles, so the piece of California, filled with solar panels, needed to generate what California consumes is 324 divided by 163,696 or .0019. This is 19/10,000ths, or about 2/1000ths or about 1/500th or 1/5th of one percent. **One fifth of one percent of California, filled with solar panels, would generate enough electricity in one year to satisfy California’s total electricity consumption for one year. Mr Sorman is wrong by a factor of 500, which is the same as being 50,000% wrong.**

While there is nothing inaccurate above, I have left things out in the interests of clarity. For example, panels are never packed so tightly together (maintenance access is needed, etc), some of California is forests or lakes or steep mountains or streets and highways and so on. Mr Sorman doesn’t mention these things either, and considering the magnitude of his error, they amount to nothing. A great source to learn more about this (and many other things) is Physics for Future Presidents by Physics Professor Richard Muller. Prof. Muller has several other books with similar titles (shown on the link above), and they are all excellent.

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