Arnold Kling  

Solar Power

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Joshua Gans writes,


What interested me more was the growing evidence that solar energy might be subject to a Moore's Law rate of improvement. According to Ramez Naam in Scientific American, the cost of solar photovoltaic models has been falling at an exponential rate since 1980. Installation costs have been falling too. So much so, in fact, that in a decade, solar would outperform the average kilowatt energy cost in the US. A decade after that and it will be approaching the cheap baseload fuels.

A slight quibble. I think of Moore's Law as saying that computer power per dollar is roughly doubling every few years. The article referred to in the quote above says,

Averaged over 30 years, the trend is for an annual 7 percent reduction in the dollars per watt of solar photovoltaic cells.

I was relieved to find that the exponential curve is not as steep as Moore's Law. I feel more comfortable with the forecast that solar will be competitive with other fuels in 2020 if it takes only an annual improvement of 7 percent to get there. If it were to require a doubling every two years until then, I would be more worried about running into impossible technological hurdles that keep solar power from ever becoming competitive.

My read on the situation is that government should still be subsidizing work at the research phase, not at the deployment phase. When solar power is cost effective, deployment will not require subsidies.


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COMMENTS (17 to date)
Mike OBrien writes:

A slight quibble - Moore's law actually says nothing about computer power per dollar. It concerns the # of transistors than can be placed inexpensively on an integrated circuit.

Julien Couvreur writes:
[...] government should still be subsidizing work at the research phase

Why?

Patrick R. Sullivan writes:

Krugman is pumping solar power, along precisely this line, in his Times column today too. Pretty suspicious.

'If you build it (with subsidies), they will sun?'

Patrick writes:

@Julien Couvreur: Positive externalities.

While all the "green" energy sources are dilute and hard to capture, I can imagine a future in which solar cells can be printed out in huge volumes for a pittance. If it's good enough to power all the plant life on Earth, it should be good enough for us.

Mark Brophy writes:

Positive externalities from solar power? That stems from the idea that solar power pollute air less than other forms of energy. But if air pollution were something that people cared about, then real estate in Los Angeles would be cheap, and areas with clean air would be expensive. In another example, Fort Collins, Colorado has clean air and cheap real estate relative to Denver, where the air is dirtier and real estate more expensive.

Patrick writes:

@Mark Brophy: No, positive externalities from the research, not deployment.

DMK writes:

Moore's Law can't budge the fact there's only so much incident energy available from the sun; 5.38 kwh/sq. meter/day, annual average, in a place like Phoenix, for example. This means the photovoltaic collector area must completely cover an area 1.3 miles x 1.3 miles to replace a typical large 1000 MW powerplant - IF the PV technology were 100% efficient! But at present 15-20% efficiencies the collector area would need to be nearly 3 miles x 3 miles!

Patrick writes:

@DMK: You're assuming total centralization. There's enough space on our rooftops, though economics could very well favor some centralization. Furthermore, solar seems more appropriate for peak-load rather than base-load. However, if someone can create cheap cells which run on infrared (heat), that could change everything. It could solve the energy storage problem.

I guess we'll know in ten years.

Arthur_500 writes:

Why should government be subsidizing research?

Government gets nothing in return from research. OK so they get a return on their top secret research for weapons, etc. but even then the private companies are doing most of the work and they are getting paid handsomely in return for the contracts.

Let's say that government got royalties off its research. Then I might be willing to back it. However, I am more certain that overall we are best served with a more limited government.

Government will do fine off the taxes from the proceeds of successful private investment of research.

matt writes:

@DMK, so with US capacity of 813,000 MW we would need an area of only 85miles squared, or about an 7% of the state of Arizona to power the entire US with no air pollution. Sounds good to me!

jb writes:

there must be something I'm missing in DMK's math.

If I assume 5kwh / square meter, it requires 200,000 square meters to generate 1000 megawatts, which is roughly a 450 meter square, which, while pretty large, is less than 1/3rd of a mile, at 100% efficiency.

What did I miss? I'm not well-versed in power-related math, so this question is in earnest, not sarcastic.

DMK writes:

Knew the math would be dangerous. First, 5.38 is an average, over a year, of daytime/darktime, sunny day/cloudy day, changes in sun's position, etc. But 5.38 kw-h is a daily figure; one average 24 hour period. For simplicity, assume sun never sets but stays in "average" position 24 hours per day. Each square meter of PV array will generate 224 watts if PV eff. is 100%. Here's a different, simpler number to remember: On a clear day with the sun directly overhead (or "solar noon" at the equator), the sun gives 1000 watts/sq. meter (1 kw/sq. meter). Note this is an instantaneous value only (kw, not kw-h). The difference between 1000 and 224 being night-time, clouds, dust, sun position, etc.

bobroberts writes:

Every square meter of earth near Phoenix receives 5.38 kWh of energy in one day from the sun. So to replace the power production of a plant that produces 1000 MW of power, take:

(1000 MW)/(5.38 kWh/(day*m^2))

to find the area needed.

1000 MW = 1E9 W
5.38 kWh/(day*m^2) = 224.167 W/m^2

(1E9 W)/(224.167 W/m^2) = 4.461E6 m^2

Then take the square root of this number to find the length of a square that will contain this area:

(4.461 m^2)^(1/2) = 2112.1 m

2112.1 m * (6.214E-4 mi/(1 m)) = 1.312 mi

This assumes 100% of that energy is convertible; in reality only 15-20% is, so multiply the quantity of energy received by the sun by 0.2 to find the actual energy obtained:

224.167 W/m^2 * 0.2 = 44.833 W/m^2

Repeat the above calculation to find the new area. I came up with:

(1E9 W)/(44.833 W/m^2) = 2.230E7 m^2
(2.230E7 m^2)^(1/2) = 4722.8 m
4722.8 m * (6.214E-4 mi/(1 m)) = 2.935 mi

Hope this helps.

DMK writes:

jb: You have an apples and oranges problem, mathematically speaking. kwh versus kw (5.38 kwh vs. 1,000,000 kw) 1,000,000 kw x 24 hours equals 24,000,000 kwh. Now you can divide 24,000,000 by 5.38.

bobroberts writes:

For some perspective, one square meter of area at a solar power plant operating at these parameters won't generate enough electricity to properly light an average, 60 W light bulb. I live in Kansas, and we use around 1000 kWh a month in electricity. It would take a plot of land at least 19 ft x 19 ft to meet our home's electricity needs if you built a solar array at these parameters.

This, however, doesn't take into account the magnitude of our power usage, which could exceed the generating abilities of the array. This means that the smallest size array I could build and still meet my NET power needs for the month could be no smaller than 19 ft x 19 ft. It would likely have to be much bigger and include some form of energy storage to power the home at times when sunlight is lower than average or at night. These two reasons are why you don't see a lot of large scale solar power plants. Even if you could build them cheaply, they don't really meet power needs very well. Peak load, like Patrick stated, might be the best viable use for them. Food for thought.

Ramez Naam writes:

Arnold writes: "My read on the situation is that government should still be subsidizing work at the research phase, not at the deployment phase."

In general I agree with this sort of sentiment.

In the specific case of solar (or wind, or any other low-carbon technology) I would amend it slightly.

Since the energy that solar displaces (coal and natural gas) produces carbon emissions that produce negative externalities, those other forms should be paying for those externalities, in a form such as a carbon tax, carbon emissions permits, or their equivalent.

In the absence of that, I think the second best option is for governments to fund deployment of low-carbon technologies in an amount proportional to the quantity of carbon emissions they displace.

As a practical matter, in any rate, boosting deployment effectively boosts R&D: It drives more revenues to solar companies and further attracts investors to solar start-ups.

And, as you say, "When solar power is cost effective, deployment will not require subsidies." That point is increasingly near.

best,
Ramez Naam

Mark Bahner writes:

Ken Zweibel

I think it was Ken Zweibel of the DOE who came to Va Tech circa 1979. My memory is (obviously) hazy about the details, but I think he was estimating that photovoltaics would become cost-effective in the mid-1980s.

P.S. Apologies to Mr. Zweibel if he instead said "circa 2020."

P.P.S. Double apologies if he wasn't the person who spoke.

[mistyped link fixed--Econlib Ed.]

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