A California-sized solar panel

Guy Sorman’s The End of Green Ideology begins:

The meltdown at the Fukushima nuclear plant has sent political aftershocks racing around the globe. More often than not, however, the shocks have been ideological, with no basis in science.

I’ve been reexamining the case for nuclear power for a while now. I still think we’ll probably need to keep developing it, improving it, and including it in the mix. But I’m willing to be convinced otherwise and I’ll watch Germany’s forthcoming nuclear exit with great interest. In any case, I agree with Guy Sorman: let’s make sure our energy future is based on science not ideology.

So I was puzzled to find this assertion in the middle of his critique of green ideology:

If California were to rely on solar power for its electricity consumption, the entire state would have to be covered with photovoltaic cells.

Really? That seems wrong. According to an infographic called Surface Area Required to Power the World (with zero carbon emissions and with solar alone), assumptions and methodology here, it looks like a California of solar panels could more than power the world.

For reasons Sorman alludes to — cost, environmental impact — it probably wouldn’t make sense to build a California of solar capacity, regardless of how it’s distributed around the world. But let’s just focus on the assertion that California would need all of itself to keep the lights on using solar photovoltaics. Is that really wrong? If so, how wrong? What’s the right answer? And how can we find out?

Before the advent of WolframAlpha I would sometimes idly speculate about these kinds of questions, but I wasn’t able to effectively analyze them. Now I can. From various sources, I think I know that a PV panel can produce on the order of 10 watts per square foot. So we can ask WA directly:

area of california in square feet * 10 watts per square foot = 45.64 TW

How can we think about 45 terawatts? I recall from Saul Griffith’s Climate Change Recalculated that the whole world runs on about 16TW. This confirms what I gleaned by eyeballing the infographic: Solar California could more than power the world.

But where does Saul’s 16 TW number come from? Can we cross-check it? The infographic assumes that we were running at 500 quadrillion BTU in 2008, projected to rise to around 680 quadrillion BTU in 2030. The US Energy Administration’s International Energy Outlook 2010 cites similar numbers. I’m not good at unit conversions, but WA is:

500 quadrillion btu in watt years / 1 year = 16.73 TW

This matches so closely that I wonder about the provenance of the number. Do all these analyses use the same source, or are there independent estimates that converge? But anyway, let’s assume that 16 TW is in the ballpark. How much electricity does California use? (Recall that this was the quantity of energy Guy Sorman thinks would require a California-sized solar panel to produce.) A couple of sources say that California uses about a quarter million gigawatt hours of electricity in a year. How much is that? WA provides useful comparisons:

a quarter million gigawatt hours = 1/15 total production of electrical energy in US in 2001 = 2 * electricity consumption of Norway in 1998 = 28 * daily electric energy production of all nuclear power plants

a quarter million gigawatt hours / 1 year 2000 hours = 28.5 125 GW

45 TW is about 1600 360 times more than that, so I think that’s how wrong Sorman’s claim is.

If a California-sized solar panel could produce 45 TW, then how much of California would be needed to produce 28.5 125 GW? Let’s ask:

area of California / 1600 360 = 102 455 square miles = 1/15 1/3 area of Rhode Island = 2.2 * area of Walt Disney World 1.1 * land area of Hong Kong

As with other essays in my energy literacy series I am mainly interested here in how tools like WolframAlpha can help us reason our way through a thicket of quantities and unit conversions. Hundreds of people liked Guy Sorman’s essay on Facebook. Dozens tweeted it and the one comment I can retrieve calls it a “balanced and scientific assessment of world’s current and future energy outlook.”

I don’t know enough to agree nor disagree with Guy Sorman’s conclusion:

Future energy supplies will most likely rely more and more on miniaturized nuclear plants and shale gas — a mix capable of responding to a rapidly urbanizing world population’s growing demand for electricity.

But I do know that we need to be able to verify and reproduce analyses that purport to be scientific. WolframAlpha has become the View Source button that helps me do that.

26 thoughts on “A California-sized solar panel

  1. Walt Crawford

    While this is a great analysis and use of Wolfram|Alpha, the latter oversimplifies one crucial aspect: Actual equivalent full-power-generation hours per year. For our part of sunny California (and our own south-facing rooftop photovoltaic system), last year, that turned out to be about 1920–which means you’d need about 130GW capacity, not 28.5, if we were typical (there’s no such thing). That doesn’t change your overall discussion…but the numbers are a little more subtle. Now: What portion of CA would need to be used for capacitors or storage batteries if you actually attempted to go all-solar?

    Reply
  2. Jon Udell Post author

    or our part of sunny California (and our own south-facing rooftop photovoltaic system), last year, that turned out to be about 1920–which means you’d need about 130GW capacity, not 28.5

    Ah, right, thanks for the reminder. The analysis at http://www.landartgenerator.org/blagi/archives/127 used 2000 hours, I forgot to include that. Fixes shown above.

    What portion of CA would need to be used for capacitors or storage batteries if you actually attempted to go all-solar?

    Oh, we’re using Nevada for that, didn’t you tell them? :-)

    Reply
  3. Jon Udell Post author

    By the way, are estimates like “modern photovoltaics (PV) solar panels will produce 8 – 10 watts per square foot” adjusted for average output over time, or should the above 45TW also be reduced accordingly?

    Reply
  4. Walt Crawford

    Thanks for the changes.

    As for adjusting the output estimate for average output over time, that’s trickier. It’s a rate claim, not a quantity claim (that is, it’s watts, not watt-hours), so that’s OK. On the other hand, solar panels do lose some efficiency over time…but that starts to get to microanalysis: The difference between, say, 455 square miles and 500 square miles (over quite a few years), not the difference between 455 square miles and California.

    Reply
  5. Pingback: Going Green Makes Sense and Cents

  6. Armin Hanisch

    Great post that shows how to use WA to shift such discussions to a more scientific side. Thanks to Walt for mentioning the useable hour span per day. Living in Germany roughly 5 miles next to two nuclear power plants (one of them will be shut down permanently now) I can assure you that we do watch the outcome of our exit strategy as interested as you, Jon ;-)
    There are some interesting calculations on the Desertec foundation’s web site also that support your estimates.

    Reply
  7. Speed

    Calculations should start with the industry standard insolation of 1,000 watts per square meter when the sun is directly overhead on a clear day. Since California is north of the tropic of cancer the sun is never directly overhead. And since the earth rotates on its axis, the sun moves from the eastern horizon to the western horizon over the course of a day and parties all night on the other side of the world.

    The most efficient solar panels claim 19% efficiency but you will pay for that. Figure 15% when brand new.

    As mentioned above, the output is a rate (watts –> power) and needs to be integrated to get energy (watt-hours etc.).

    This would be an interesting assignment or class project for a high school AP math class. I expect that Wolfram Alpha can handle it if you can figure out how to present the problem. Don’t forget to figure in the average cloud cover.

    Reply
  8. Ron Swenson

    I’ve done a lot of analysis along these lines and with the corrections suggested, you’re in the ball park on the supply side. It can be tweaked, for example, by considering where you are installing solar in California. On the coast for PV you will see about 4 kWh/DC-rated-kW/day; in the Mojave, 6.5 or more. With solar thermal electric it could be more.

    Missing from the analysis is the emerging technology changing on the demand side. As we adapt to solar, for example, in transport, we can move our vehicles above the organic plane, save (worldwide) most of the million lives lost in traffic accidents and countless injuries annually, while cutting energy consumption by a factor of 5 or more. You can see this at http://www.solarevolution.com.

    Anyone suggesting natural gas for transport is clearly not thinking for the benefit of our children. Life requires energy; fossil energy is finite; life based on fossil energy would be finite.

    Reply
    1. Speed

      Nevermind. I just figured out your units.

      Assuming 15% efficiency, a 6.67 square meter panel would be rated at 1kW. And you calculate an average output of 4kWh per day for such a panel on the coast and 6.5 kWh per day in the Mohave.

      Reply
  9. Bender

    If you’d like to learn more about shale gas as mentioned in Guy Sorman’s conclusion, google for ‘fracking’. Even better, google for a documentary called ‘Gasland’. It is definitely worth your time to do so.

    I like the idea of modern nuclear plants, but I don’t believe shale gas should be an option.

    Reply
  10. Pingback: Daily Digest for July 13th

  11. Jon Udell Post author

    Wow. Thanks James! I love the phrase “active reading” and the implementation — with not only View Source but also Modify Source — is terrific!

    Reply
  12. Michael Große

    For unit conversions and big picture overview about what energy is needed and what can be generated in the UK look at Sustainable Energy – without the hot air from a Cambridge physicas professor. I would hope something like this existed for every major country…

    http://www.withouthotair.com/

    Reply
  13. search engine optimisation newcastle

    It’s funny i stumbled onto this post. I have recently got a solar panel quote for my house in Australia. Going to cost 18k for 21 panels. It will completely offset my power bill though. And with the rebates we are currently getting it makes it hard to say whether it is worth it.

    Reply
  14. Pingback: Photoshop

  15. Pingback: Solar Sun Rings – Solar Pool Heater - Swimming Pool Pumps and Filters | Best Swimming Pool Pumps and Filters | Best Swimming Pool Pumps Reviews

  16. Linda Mcmahon

    “The assumptions Udell makes assumes current solar panel efficiencies. Combine advances in panel manufacturing techniques with the continued decline of solar panel prices a California sized solar panel will be able to support the increased energy demands of 2050.”- Will this happen?

    Reply
  17. Pingback: diy solar

  18. Stupid

    Your wrong technology at this time would not support it 20% need to average factor in weather heat decreases efficiency energy for producing highly pure silicon 99.99999%. Damage to habitats water to produce panels. Politicians are always right no matter the garbage out of their mouths all the radio talk jockys repeat it then it must be true. When people think they have all the energy they could possibly want. The n they commsume %250 more. Then they develop electric cars which then drive %350 more which means they comsume %500 more junk then they build more power plants to meet demand then the vicious cycle repeats but with bigger numbeers

    Reply

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s