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Final Comments on Solar Posts

I am going to be offline for a few more days, enjoying some time with the family. In the interim, Tom Standing has sent some detailed replies to some of the comments following his posts Arizona Solar Power Project and Ambitious Solar Plans in France.


Here is some additional material in response to a few of the comments that were submitted regarding my essays on the solar project in Arizona and the solar plan for France.

First is a general comment about my intent with the two essays. I am merely attempting to contribute some hard-edged reality to many solar proposals that do not seem to have been adequately appraised through the conceptual engineering process. The value and scale of proposals for renewable energy projects will be demonstrated through the laws of conservation of energy and thermodynamics. Rational calculations employing these laws are what I am basing my analysis on. We have at our disposal an immense database of solar insolation data on the NREL website, from which we can estimate how much energy a solar project is capable of delivering, and how the energy would be distributed with respect to time. I have also attempted to describe reasonable assumptions to fill in gaps of data or other information in order to complete the calculations. I fully realize that reasonable people will disagree with some of my assumptions, but I think that the differences will not significantly alter the conclusions.

Someone offered energy consumption by a range of vehicle size: one megajoule per mile at highway speeds for light vehicles, 2 MJ for heavier vehicles, and 10 MJ for 18-wheelers.

May I suggest we convert these numbers into units that we are familiar with, such as miles per gallon? A few key conversion factors can be used, as follows.

  • 1 joule/sec is the definition of 1 watt; therefore one kilowatt-hour is 3.6 million joules.
  • The heat equivalent of electrical energy is about 3,535 Btu per kWh (100% conversion).
  • Therefore, 1 Btu = 1,020 joules.
  • The heat value in 1 U.S. gallon of gasoline is 125,000 Btu.
  • The heat value in 1 U.S. gallon of diesel fuel is 138,000 Btu.

Working out the numbers, 1 MJ = ~ 1,000 Btu, which is 1/125 gallon of gasoline, which, according to the original comment, light vehicles could travel 125 miles/gal.

At 2 MJ/mile, SUVs would travel ~ 62 miles/gal.

The 10 MJ per mile for 18-wheelers burning diesel fuel calculates out to 13.8 miles/gal. Are these reasonable consumption rates? Most people would expect vehicular fuel consumption to be substantially higher.

The same commenter opined: “if a one square meter heterojunction panel can squeeze out 1.6 kWh = 5 MJ a day…”

Let’s estimate what the conversion of insolation to useful electricity would be for this panel, using NREL insolation data.

California’s Mojave Desert offers the highest annual average insolation of any of the 239 monitored stations in the U.S. – 6.6 kWh/ (m2-day) for unshaded fixed panels facing south, tilted at an angle = local latitude. If the panel yields 1.6 kWh of useful electricity in a day, then the conversion factor = 1.6/6.6 = 24.2%. If that same panel were exposed to insolation in St. Louis, MO where, with the same panel orientation, average annual insolation is 4.8 kWh/ (m2-day), and yields 1.6 kWh/day, the conversion factor = 1.6/4.8 = 33.3%. That would be a pretty good panel!


Here is my response to another comment about the French solar plan. The comment was by Bob Lynch, December 22 at 7:43 PM. He wrote: “The French though DO HAVE areas that bask (Basque?) in cloudless days, week after week. France has a strong infrastructure to deliver power far from where it is generated, so it does not have to be on millions of 17th century homes and hovels that dot the countryside.”

This, then, is my response:

If I interpret Mr. Lynch’s vision accurately, he believes that France could construct a major portion of their solar arrays in their sunniest regions, and then transmit the generated electricity to the populous regions where the climate is less favorable to collecting solar energy.

We can check the validity of such a proposal with the insolation data posted on the NREL website.


Click the link “In alphabetical order by state and city.” Then choose any city to obtain the data in spreadsheet format.

The insolation statistics I use here are for flat-plate collectors facing south at a fixed-tilt angle equal to the latitude of the site. This orientation gives the optimum solar exposure for fixed flat plates. Most installations will not match this ideal orientation. Collectors are tilted at varying angles, may not face due south, are not always clean, and may experience shading from nearby buildings or trees. Generally, solar installations generate about 15% less electricity than is calculated from insolation data and the manufacturer’s conversion factor.

Although the NREL data covers 239 stations in the United States, we can closely approximate insolation in France with comparable locations in the U.S. based on equivalent latitude and similarities in climate. The southern-most part of France, which provides the highest insolation, is in the range of latitude 43 to 44 degrees. A comparable location where the climate features “cloudless days, week after week,” is Boise, Idaho, latitude 43.57 degrees, with a semi-arid climate. The NREL data shows powerful insolation during the 6 months April through September of 5.8, 6.2, 6.5, 7.0, 6.8, and 6.5, respectively, in units of kWh/ (m2-day). The 30-year annual average is 5.1 (same units). I think we can say that there are scant areas in France where insolation would exceed that of Boise.

The insolation that I estimated as an average for France is 4.6, which, I think, is a fair representation of the regions where solar is most apt to be developed.

An important fact that we need to keep in mind is that average annual insolation does not vary greatly over wide reaches of the U.S. Similarly, variations in Europe would be narrow. For example, Sioux Falls, South Dakota is 850 miles due east of Boise (identical latitude), but with a humid continental climate. However, the 30-year average insolation is 4.8 (same units). It’s a good bet that across the southern quarter of France, insolation would be in the 4.8 to 5.1 range, hardly a bonanza for solar development.

Some 260 miles north-northwest of Boise is Spokane, Washington, latitude 47.63, the latitude that is about 80 miles south of Paris. Summer insolation in Spokane is generous, but noticeably below that of Boise: the 6 months April – September: 5.2, 5.6, 5.9, 6.5, 6.3, and 5.7, respectively. The 30-year annual average is 4.5. Spokane’s insolation is, therefore, likely to be near that of Paris and across the northern third of France. Thus, the 11 or 12% difference of insolation in France, between the sunniest south and the north, is probably not sufficient to justify transmission of large quantities of electricity. Solar-generated electricity would best be utilized near the source.

December 29, 2008 Posted by | Arizona, France, reader submission, solar PV, solar thermal | 76 Comments

Ambitious Solar Plans in France; Solar Capacity Factors

The following guest post was written by Tom Standing, a “semi-retired, part-time civil engineer for the City of San Francisco.” In Part 1, Tom took a critical look at a 280 MW solar thermal plant in Arizona. Here in Part 2, Tom examines France’s ambitious solar plans.


The December 1 issue of the Oil and Gas Journal carried a “Quick Take” article about France’s “national plan for renewable energies” that they unveiled on November 17. Their plan includes all the popular ideas for alternative energy: biomass, wind, hydro, waves and tides, with a major emphasis on solar. For now France has 13 megawatts of installed capacity in solar, but the energy minister wants solar to be a whopping 5,400 MW in 2020! He says that France will change its carbon-based energy model to a completely decarbonized model: each home, company, and community will produce its own energy.

Excuse me, but why is France doing this? They already have the least carbon-intensive energy system of any industrialized nation. They generate 75% of their electricity with nuclear, supported by the most extensive technology to reprocess spent nuclear fuel, than any nation in the world. Practically 100% of their rail system is electrified, packed with people, whether on the Paris Metro or speedy intercity trains. France has already developed the working model of a low-carbon energy system for other nations to emulate.

Let’s do some rational calculations on France’s solar plan, similar to my last email. We can see what surface area of collectors would be needed, and how much electricity would be generated.

Collector Area

As I explained previously, solar panels are rated at their maximum output, when the sun is near its highest altitude for the year under cloudless skies. Under such ideal conditions, insolation is about 1,000 watts per square meter. The most cost-effective panels convert about 10% of insolation into useful electricity, a factor that has remained unchanged for 10 years. Some PVs might convert 15%, but they cost more and are not mass-produced. Thus a typical panel of one square meter is rated at 100 watts.

To estimate the area of PV panels that France wants to install, we simply divide 5,400 MW by 100 W/m2 and we get an incomprehensible 54 million m2! It means that one million homes and businesses would have to be covered with 54 m2 of panels. A typical home can accommodate only 25-30 m2, so more than a million buildings would have to install PV. I do not know the worldwide capacity for manufacturing PV panels, but I would guess that current capacity is a small fraction of 54 million m2/year. Oh sure, capacity will grow, but what about PV for Germany, Spain, UK, the Low Countries, and the US? California alone would suck up a major chunk of that capacity.

Solar Electricity Generated

Let’s say that France actually installs 54 million m2 of solar by 2020. (I think it’s fantasy in the extreme, but let’s carry the scenario through.) How much electricity will the fully built-out system generate?

First, we need to estimate the insolation upon the collectors. While I have copious insolation data from NREL for the US, I have no site-specific data for Europe. But I can make a reasonable estimate. Having traveled throughout France at various times for a total of about 3 months (typically in the summer), I can say that France has mild sun conditions. I would compare the French summer sun to that of Cleveland or Minneapolis. However, France is located at somewhat higher latitudes, which tends to reduce midday sun strength and spreads it out over more daylight hours. The northern suburbs of Paris, say around de Gaulle International, is latitude 49o, the northern-most boundary between the US and Canada. The south-most reaches of France, are between latitude 43 and 44, equivalent to Buffalo, NY or Portland, Maine.

I will pick a number on the generous side for annual average insolation in France, equivalent to that of Boston, New York, Chicago, and Minneapolis:

4.6 kWh/ (m2-day).

This level of insolation is for optimum panel orientation: facing due south with no shading, tilted at an angle equal to local latitude. Varying amounts of shading with less than ideal orientation will reduce the insolation on the collectors of most installations.

Now we are ready to calculate the annual energy generated from the fully-built French PV system. As I showed in Section 8 of my previous email, the annual energy generated by a solar installation is the product of four factors:

Insolation, average day during a year = 4.6 kWh/ (m2-day)

10% conversion of insolation into electricity, the industry standard for PV

Area of solar collectors = 54 million m2

365 days/year

Cancelling out units and carefully watching orders of magnitude, we come up with 9 billion kWh of useful electricity generated during the first year of complete build-out. But we need to give this number some perspective.

Energy Generation in Perspective

EIA statistics show that French consumption of electricity grew from 353 billion kWh in 1992 to 415 billion in 2002, or 62 billion kWh in 10 years for an average gain of 6.2 billion kWh/year.

It means then, that this huge solar development would, at best, produce the equivalent of only 1.5 years of gain in France’s electricity consumption. And it would take a 12-year crash program to install that much solar!

Another comparison is with the annual power output of one of France’s 1,000 MW nuclear power reactors. If the reactor operates for a year at 90% capacity (typical for the industry), the three factors to multiply are: 90%, one million kW, and 8,760 hours/year. Multiplying out these factors, we find that a single reactor would produce about 8 billion kWh/year, roughly the equivalent electricity as all the solar panels covering nearly 2 million homes and businesses.

Costs for PV

In the US, homes and businesses that install PV typically receive “rebates,” (another word for “subsidies”) from state or local governments, or the utility, to be paid for by all ratepayers. Rebates usually amount to about half of the total installed cost. The unit cost for solar installations has changed little since 2000, in the range of $600 – $700 per square meter, or in the terms of the industry, $6,000 – $7,000 per rated kW. Thus a homeowner usually qualifies for a rebate of 6 or 7 thousand $ after installing a 2 kW PV system.

I don’t know if French taxpayers and ratepayers will subsidize solar installations, but the unsubsidized cost remains the same and must be paid by somebody. Total installed cost of the solar plan for France, then, would run in the neighborhood of $35 billion. What is the cost of building a single reactor in a nuclear power plant? Considerably less I would say.

Solar Capacity Factors

We can calculate capacity factors for any solar project directly from insolation data. This provides a shortcut to calculating electrical output on an annual or monthly basis when we are given the nameplate capacity. The capacity factor depends only on insolation during the time period in question, and is independent of the conversion factor between insolation and electricity.

The capacity factor is defined as the ratio of actual energy generated, to the energy generated at maximum insolation, i.e. nameplate capacity. Therefore, our ratio is:

actual average isolation/maximum insolation

Maximum insolation, we have seen, is 1,000 W/m2.

Actual insolation is given in kWh/ (m2-day). We have to convert this quantity into units of W/m2. We cancel out hours and days by dividing actual insolation by 24 h/d. For example, if insolation is 5.0 kWh/ (m2-day), the average power output during one day is 5,000/24 = 208 watts. Average power output divided by maximum insolation (1,000 W/m2) gives a capacity factor of 20.8%.

If you have some time during this busy season to look at this, I would value your input. I would like to take the discussion about “green energy” beyond all the vague generalizations that we hear repeatedly from our leaders about stimulating a green or clean energy future. It’s as if “renewable energy” is a virgin topic, yet to be assessed and waiting to be tapped. But we already have a great amount of information with which to evaluate avant-garde energy proposals. We need to put hard numbers to these generalizations.

December 22, 2008 Posted by | France, reader submission, solar PV, solar thermal | 55 Comments