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German Robot Pigs

“What I need for this project”, I explained to my young son, “is an army of German robot pigs.” His eyes went wide. “And I know just where to get them…”

I have a project on my desk right now to look at avenues of disposal for sewage sludge in a specific location. There are lots of things you can do with sewage sludge, but it is obviously dependent on a number of factors. There are places where farmers spread it on their fields, there are places where it is incinerated, and there are places that it is land-filled. And there are other options beyond that.

In order to utilize sludge for energy, the water content is obviously critical. Too much water, and any energy you create from the sludge is lost due to the need to dry the sludge. So what I really needed in this specific case was a method of drying that could help generate net energy from the project.

Obviously you could just spread the sludge out in a sunny location, but there are many problems with that. One good rainstorm can ruin a week’s worth of drying and create a brown river to the nearest waterway. Further, as the outside layers dry, the drying slows down unless the sludge is constantly mixed.

I thought that this particular situation could benefit from a solar dryer. I am familiar with solar kilns and solar ovens, so why not a solar sludge dryer? I envisioned a system like a greenhouse that allows the sludge to dry, but also ventilates and removes the moisture. Several of the functions would need to be automated if this has any hope of being economical.

So I thought “Hmm, I wonder if anyone has invented such a system.” Since about 98% of anything that I ever think of has already been invented, I Googled it. Turns out that there are lots of solar sludge dryers to choose from, but one in particular caught my eye. Enter the German robot pig:

German robot pig the star of sludge drying plants

A group of young, up-and-coming German scientists in the late 1990s founded a company that today is the world leader in solar sludge drying. The star of the environmentally friendly drying process is an unusual pig that works all day without food or complaint.

What would a machine look like if it lived in the mud? Well, probably like a pig. Some 12 years ago, Tilo Conrad, together with two of his fellow students from the University of Hohenheim, built the first electrical pig, pioneering a device that is now being used as a solution to waste disposal problems throughout the world.

The stainless steel pigs, which in a sense resemble big beetles, are an important part of a larger solar drying process was patented by Thermo-System GmbH, a company Conrad founded in 1997 in Filderstadt.

Today, nearly 200 mechanical pigs do what normal pigs do: wallow in and shuffle through the mud — only these pigs do it to reduce sewer sludge disposal costs and protect the environment. Conventional drying processes use non-renewable energy, but Thermo-System’s process harvests solar power to dry the mud. It spreads the wet sewage sludge into sheds that are similar to greenhouses and puts the pigs to work.

In the sheds, the sludge absorbs heat from the solar rays and an innovative ventilation system keeps the air inside the shed warm and dry. The electrical pig, which is a fully automated robot complete with stainless steel mixing tools, tills and aerates the microbiologically active sludge, thereby accelerating the drying process. The whole system is fully automatic, uses very little energy and can be easily maintained.

Alas, another invention that someone thought of 10 years before I did. But this system really has many of the characteristics that I am looking for. Whether the economics pan out is still entirely unclear. The company does have an impressive project portfolio, though, having already built over 100 sludge-drying plants worldwide.

But one doesn’t get a chance to work with an army of German robot pigs every day, so I will continue to investigate and maybe this works out. Besides, that sounds so much more interesting than “I am working on sewage sludge.”

December 20, 2009 Posted by | solar drying, solar thermal | 18 Comments

Running the Electric Grid with eSolar

As I often do on a Saturday morning, I was up early reading through energy headlines. I happened across this story on eSolar:

Bill Gross’s Solar Breakthrough

“We are producing the lowest cost solar electrons in the history of the world,” Bill Gross is telling me. “Nobody’s ever done it. Nobody’s close.”

“We have a cost-effective, no-subsidy solar power solution and it’s for sale, anywhere around the world,” he says.

The article was intriguing, and inevitably led me back to eSolar’s website to get a better idea of whether the claims appear to have merit. There, I watched the slide show on the technology, and caught this bit: A single unit generates 46 MW of clean electricity on a footprint of 160 acres.

While this doesn’t help me figure out whether they can deliver on the hype, it does enable me to update a couple of essays that I have written before:

A Solar Thought Experiment

Replacing Gasoline with Solar Power

In the first, I made an attempt to calculate the area that would be required to equal the entire installed electric capacity of the U.S. – using only solar power. (Yes, I understand that this number falls to zero at night). The numbers quoted above from eSolar – combined with the latest data on installed electrical generating capacity – enabled me to update that calculation.

Per the EIA, total installed electrical generating capacity in the U.S. is approximately 1 million megawatts. If we scale up eSolar’s claim of a required footprint of 160 acres to produce 46 MW of electricity, then it would require 5,435 square miles of eSolar technology to equal current U.S. electrical capacity. This is a square of 73.7 miles by 73.7 miles. This is greater than the 2,531 square miles calculated in the previous essay, but that essay only considered the area for solar panels. The present calculation encompasses the footprint of the plant.

Looking back at the gasoline calculation, I came up with 1,300 square miles required in my previous essay to replace the energy gasoline provides. Using the current eSolar numbers changes that number to 2,413 square miles, or a square of 49 miles on each side.

Of course all of the normal caveats apply as spelled out in the previous essays. The key point is not to read these sorts of thought experiments too literally. I tend to do them to get my head around the scale of certain problems. Complaints of “the cost is too great” or “the power is intermittent” – addressed by caveats in the previous essays – completely miss the point of the essay. It is sort of like trying to figure out how much biomass would be required to power the world. If the calculation is 10 times the current annual output of biomass, then that’s not going to work. If it is 1/100th the current annual output of biomass, then that might work (again, pending lots of other things working out).

In this case, I find this eSolar thought experiment encouraging insofar as the required land area isn’t a clear knockout.

August 8, 2009 Posted by | electricity, electricity usage, eSolar, solar efficiency, solar power, solar thermal | 29 Comments

Solar Stories

A couple of interesting solar stories this morning, as well as a new blog covering solar power. First, the new solar-focused blog by Paul Symanski. Paul has experience in the solar industry, and many of his early entries are concerned with solar energy economics:

Rate Crimes – Bringing Transparency to the Economics of Solar Energy

From Paul’s first entry in May – There is No More Important Energy – he writes:

The Rate Crimes conversation centers on solar electric energy because of its importance to the future of our society: a society that is defined by electric energy as much as by the fuels that currently provide us mobility.

Solar electric energy has myriad advantages over the traditional fuels that provide us with electricity. Solar energy is plentiful, clean, immediate, proximate, distributed, mobile, scalable, unobtrusive, long-lived, durable, gathered, simple, safe, unassailable, independent, equitable, and profitable. And, like no other energy source, solar energy has the potential to become ubiquitous.

Solar energy is plentiful. Enough solar energy falls on the Earth in one hour to power the whole planet for an entire year. Resources for exothermic reactions (e.g. combustion, fission) diminish. As this occurs, these traditional fuel resources will no longer be able to meet our demand for energy. Energy generated by the photoelectric effect will supplant the traditional fuels.

Next, a pair of headline stories this morning about solar power:

Nation’s largest solar plant to be built in NM

ALBUQUERQUE, N.M. (AP) — Utility officials announced plans Thursday to build a giant solar energy plant in the New Mexico desert in what is believed to be the largest such project in the nation.

The 92-megawatt solar thermal plant could produce enough electricity to power 74,000 homes, far exceeding the size of other solar plants in the United States. The largest solar thermal plant in operation now is about 70 megawatts, said Dave Knox, a spokesman for New Jersey-based NRG Energy, the company building and running the facility.

“This is larger than anything in existence in America so far today,” he said.

It will be similar in many respects to a steam plant, using the sun instead of fossil fuel to generate steam and produce electricity, said Michael Liebelson, president of NRG and chief of development for its low-carbon technologies.

I have been thinking a little about the intermittency issue. I wonder if you could have a natural gas tie-in, and whenever your thermal mass starts to cool off after the sun goes down, just keep it heated up with natural gas. I haven’t heard of this being incorporated into these solar thermal plants (although maybe it is?), but it seems to make sense to me. The capital costs would be higher, but you then have a plant that can run 24 hours a day – with solar contributing perhaps 2/3rds of the power. Of course if you have enough thermal mass, you could potentially keep the plant running overnight anyway before things cooled off to the point that you can no longer produce electricity.

[Note: A reader sent me a link to show that yes, someone has started to build a hybrid plant incorporating the elements I mentioned above: FPL Breaks Ground on First Hybrid Solar Plant]

The second story is from India:

India plans much solar power, slower emissions rise

India is about to publish eight climate “missions” to boost efficiency, renewable energy and sustainable development. “We hope that will be completed in the next few weeks,” said [Shyam] Saran [RR: Saran is special the climate envoy to Prime Minster Manmohan Singh]. One policy aim is to install about 20 gigawatts of solar power by 2020, he told Reuters.

“It’s around 20 gigawatts, that’s something we’ve been talking about.”

The world now produces about 14 gigawatts (GW) of solar power, about half of it added last year. Analysts said they want details of the Indian plan before hailing what would be a big lift to a small but burgeoning market.

Regular readers know that I am bullish on solar power in the long run. I think our long-term future will consist of electricity produced from solar, wind, geothermal, and nuclear (it is going to be a while before coal usage is substantially impacted) and liquid fuels produced from gasification and hydrocracked lipids. Even if we see lots of electric cars hitting the roads, we are going to continue to need liquid fuels for the airline industry and for long-haul trucking. Short term (say, the next 20 or 30 years) I still think fossil fuels will be our primary source of energy.

June 12, 2009 Posted by | solar power, solar thermal | 26 Comments

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

Arizona Solar Power Project, Calculations

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 takes on the calculations for a 280 MW solar thermal plant in Arizona that I looked at back in February. My conclusion from that essay was that the electrical demands of the U.S. could in theory be met on 10,000 square miles of land. Tom peels the onion a few more layers and puts the energy production into perspective.

While solar calculations are by no means second nature to me, I see no obvious errors in Tom’s calculations. But I consider peer review to be a very useful component of my blog, and I know that Tom would appreciate any constructive criticisms. Part II will delve into France’s solar ambitions.


Hello Robert,

You and I met at the Sacramento Peak Oil conference. Your presentations and discussions were most enlightening. I was heartened by your analysis of cellulosic ethanol. I have always been deeply skeptical of the notion that the U.S. might displace a meaningful portion of transportation fuel with biofuels from cellulose. I could give you some of my thoughts on this subject, but you have already covered the territory thoroughly.

I want to comment on your calculations you posted in TOD in February, regarding the proposed 280 MW solar thermal plant in Arizona. First, a bit about my background. I started my career as a chemical engineer, first in refinery operations, and then chemical processing design. But that was only about 4 years. Most of my career has been as a registered civil engineer in a variety of disciplines for the City and County of San Francisco. Over the years, I have become interested, maybe even fascinated about the prospect of utility-scale generation of electricity from qualified renewable sources.

Throughout North America and Europe, many people have focused on renewable energy as a means of reducing dependence on Middle East oil and reducing CO2 emissions. They see renewable energy as an important element to achieve emissions targets of the Kyoto Protocol. In the U.S., renewable energy from wind, solar, and biofuels appears to be a keystone for energy policy in the Obama Administration. In Texas, T. Boone Pickens is campaigning for a new American energy policy centered on major input from wind-generated energy to displace electricity generated from gas-fired power plants. Natural gas would then be redirected as CNG to power autos and trucks. In California, Governor Schwarzenegger sees his “Million Solar Roofs” program as leading other states to do likewise, thereby reducing CO2 emissions. California utilities are mandated to supply 20% of electricity from qualified renewable sources (wind, solar, bio/waste, geothermal, and small hydro) by 2015. Contributions from these sources have been stuck in the range of 10-11% since 2000. The 20% mandate appears to be a major challenge, maybe unrealistic.

Many questions come to mind in looking at the proposed Arizona plant. What precisely does the 280 MW refer to? Is it the plant’s output at capacity? Is it an annual average output? How much electricity will it generate annually? How will output vary during the day, or by season? How will output be affected by clouds?

There is important data available and a few fundamental design features that will answer these questions. Costs for construction, however, are not my strong suit. Other analysts will have much better information on costs. Cost of the plant will not change the results of my analysis.

1. Insolation Data

Reliable data for site-specific solar radiation (insolation) is critical to estimating solar capabilities. Fortunately, a massive database for insolation is posted on the National Renewable Energy Laboratory (NREL) website. In 2000, an engineer who designed solar facilities directed me to the site; I was utterly amazed at what was there. I had to be extremely selective to get the most useful data. I settled on 30-year (1961-1990) average insolation for 239 U.S. cities: monthly and annual average insolation in kWh per square meter per day. Readings for all 239 stations are given for all possible orientations of solar collectors, either fixed or tracking systems. Amazingly, insolation data is also tabulated for averages of each hour during the 30 years for all 239 stations (kW/m2), enough data to make your head spin! Data is also tabulated for insolation of all collector orientations at 239 locations above Earth’s atmosphere! For reference, I eventually copied pages that filled a binder weighing 10 lbs.

2. Site Coverage with Solar Collectors

A rough approximation for coverage of the 1,900-acre site with solar collectors is 50%. Space is needed for maintenance and control centers, electrical converter units, towers for power lines, and maybe a backup power facility fired by gas or oil. Proposed facilities to store electricity for release at night will also consume land.

In 2001, I toured a solar thermal plant at Kramer Junction in California’s Mojave Desert.


At one square mile, it is about 1/3 the size of the Arizona plant. I would say that close to half of the site is taken up by gravel roads for maintenance vehicles. At least weekly, wash trucks at night clean the collectors of dust that frequently blows around. The roads also provide necessary space between rows of collectors to prevent shading. Collectors tilted upward to gather more sunlight cast shadows at low sun angles. If the designers in Arizona are really stingy with land use, they may be able to cover 50% of the site with collectors, including facilities for power storage.

As with Kramer Junction, the entire site will be dedicated to industrial use, fenced off and completely secure. Areas covered by collectors are denuded of vegetation, graded, and compacted. There is hardly space for a rodent or a bird to live. Collectors are supported by steel columns embedded in reinforced concrete foundations designed to resist maximum wind forces upon the considerable surfaces of the collectors. These are real-world features that solar advocates overlook when they envision hundreds of square miles devoted to solar power.

3. Calculate Collector Area

We calculate the area of solar collectors in square meters to utilize NREL insolation data.

The 1,900 acres converts to 7.7 million sq m. With 50% for collectors, 3.8 million sq m are on the site.

4. Model the Collection Array

The Arizona plant is to be a concentrating system that tracks the sun. Surprisingly, NREL data shows that concentrating systems collect less sunlight per sq m than systems consisting of flat plates, one-axis tracking, tilted at an angle = to latitude of site. Thus to be generous, I will calculate the output based on flat plates, 1-axis tracking, tilt = latitude.

For Phoenix, NREL data gives average annual insolation for our model as 8.6 kWh per sq m per day (i.e. all days averaged for 30 years). For Tucson, insolation under our model is 8.7, with slight differences for each month.

5. Calculate Insolation Striking the Collectors

Here we convert solar energy striking collectors during one day, to the average rate during the day. Thus, for Phoenix (the nearest station with NREL data to the plant):

The average annual rate of solar power striking collectors

= [8.6 kWh/ (m2-day)] [one day/24 h]

= 358 watts/m2, say 360 watts/m2

Scaling up this power for the entire plant, average daily solar power striking all collectors

= (360 W/m2) (3.8 million m2)

= 1,370 megawatts

6. Assume 15% Conversion of Insolation to Useful Electricity

The solar thermal plant at Kramer Junction converts about 15% of insolation striking the collectors into electricity. Therefore, a decent assumption for the Arizona plant that would be consistent with our other assumptions is 15% conversion.

Average electrical power generated by the plant over the entire year

= (0.15) (1,370 MW)

= 205 MW

This power output is, of course, highly variable, depending on time of day, season, and cloud cover. To get an idea for seasonal changes, the NREL data tells us that plant output would average 257 MW for an average day in June, to 138 MW for an average day in December.

7. Maximum Electrical Power Output

What might be the maximum electrical power output of the plant? It would correspond to maximum insolation, which is roughly 1,000 watts/m2. Fifteen percent conversion gives a plant output of 150 W/m2, times 3.8 million m2, so maximum electricity generation = 570 MW.

According to NREL data for the desert, maximum insolation duration is about two hours a day under cloudless skies from late spring through early summer. The duration of maximum shortens with increasing time away from June 21. In early spring and late summer, maximum insolation slips below 1,000 W/m2.

Clouds have a widely variable effect, from a 10 or 15% reduction from thin cirrus clouds, to a 50-70% reduction from dense cumulus clouds (thunderheads). At Kramer Junction, operators adjust flows of the heat transfer fluid whenever a cloud drifts over the array. I seem to remember that operators engage small electric pumps to keep the fluid flowing in portions of the array that experience cooling. The Arizona array, with three times more area, will experience more frequent effects of cloud shadows.

8. Annual Energy Generated

One final simple calculation gives us the average annual electrical energy that the Arizona plant will generate. It is the product of four factors:

Insolation, average day (NREL data) = 8.6 kWh/ (m2-day)

15% conversion of insolation to electricity

Area of solar collectors = 3.8 million m2

365 days/year

Thus the 1,900-acre Arizona plant will generate roughly 1.8 billion kWh per year.

Let’s give this quantity some perspective. EIA statistics for renewable energy in 2007 show that wind-generated energy in Texas was 8.1 billion kWh. Thus it would take four and one-half plants the size of the Arizona plant to match Texas wind energy for 2007.

A more telling comparison is with the recent growth of electrical consumption in the U.S. EIA statistics show that the U.S. consumed 2,885 billion kWh of electricity in 1992; in 2002 consumption was 3,660 kWh. Average growth, then, was 77 billion kWh per year over the 10 years. Thus the electrical energy that would be generated by the Arizona plant would supply only 2.3% (1.8/77) of one year’s growth of U.S. electrical consumption. I do not have electrical consumption broken down by state, but I would guess that Arizona could build a solar plant of equal size every year, and they would barely cover their own growth in electrical consumption.

PV Potential

I have not touched on PV, but there is much to discuss. NREL data is so extensive that there is almost no limit to analyses that could be done. For now, I should only refer you to an article that I published in the Oil and Gas Journal, June 25, 2001 issue. I graphically displayed annual insolation curves for a wide range of locations. At a glance the reader can see how insolation varies with latitude, longitude, and collector orientations. I also ran through sample calculations to see how much energy can be generated. An important finding is that insolation for most of the eastern half of the U.S. stays within a narrow range: 4.6 to 5.2 kWh/ (m2-day), with fixed collectors facing south, tilted at latitude for maximal exposure.

The above calculations are purely rational, using insolation data and general assumptions in design. Actual practice shows that solar installations typically generate 10 to 15% less energy than what the calculations show.

December 20, 2008 Posted by | Arizona, reader submission, solar power, solar thermal | 36 Comments

The Solar Thermal Option

I apologize for being out of pocket lately, and that trend is going to continue at least through this week. I have a staff meeting all week, and then I fly back to Europe next Monday. So, my posting will be sporadic until then. I do appreciate everyone keeping the comments civil in my absence, as it makes for much more productive discussions.

However, I want to call your attention to a new website that discussed solar thermal in depth. The site just went live, and the topic is covered in detail. The site is:


I have always found the prospect of solar energy very attractive. In fact, I once accidentally started a fire while playing around with solar energy. However, I must admit to being surprised that the recent IEA report projected that both solar PV and solar thermal technologies will still be very expensive relative to other renewable electricity options in 2030. (See the figure that I posted here). I certainly know that each time I look into getting a solar hot water heater (which really appeals to my environmental side), my inner tightwad balks at the cost.

Car Update

And speaking of inner tightwad, I still haven’t bought a car. I have been without one since March, only renting one for a week or two at a time while in Dallas. After weighing all of the various suggestions, I had narrowed my choice down to a small truck like a Ford Ranger (which I have owned before) or a Toyota Tacoma. I don’t need to haul stuff very often, but two or three times a year I find myself needing a little truck.

But I have run the numbers several times, and I keep coming up with a cost of ownership in the $500 month range (includes taxes, insurance, maintenance, and depreciation). For the month of December, I only need a car for 4 days (this week in fact) and it will cost me less than $150 (and my insurance company covers me for no extra cost). Until I am spending more than two weeks a month in the U.S. on a regular basis, I think I can justify continuing to rent.

December 2, 2008 Posted by | cars, solar thermal | 90 Comments

Massive Solar PV Deal

Sorry for the lack of postings/comments in the past few days. I have been traveling, and just arrived back in the U.S. I have lots of things to catch up on (family time, among the most important) and I have to prepare my presentations for the ASPO conference (I will be presenting on “Biofuels” and on “Tracking Public Data” such as EIA and IEA numbers). I also have a staff meeting the first week of September. My posting frequency in the next 30 days or so is likely to drop down to 1 or 2 a week, and then should go back to normal following ASPO which begins September 21st.

In the meantime, there was a major energy story yesterday that is worth noting, and it generated some numbers for crunching. Pacific Gas & Electric in California has signed an 800 megawatt solar deal with OptiSolar and SunPower. CNET has some of the details:

Massive solar photovoltaic plants are California-bound

SAN FRANCISCO–Pacific Gas & Electric has inked deals with OptiSolar and SunPower to establish 800 megawatts of solar farms in California, which could become the world’s largest set of grid-tied photovoltaic installations.

The new plants would provide 1.65 billion kilowatt hours each year, enough to serve nearly 250,000 homes, according to Jack Keenan, CEO and senior vice president of PG&E.

“This commitment not only moves us forward in meeting our renewable goal, it’s also a significant step forward in the renewable energy sector,” he said. “Utility-scale deployment of PV (photovoltaic) technology may well become cost competitive with other forms of renewable energy generation, such as solar thermal and wind.”

The supply breaks down as follows:

OptiSolar’s 550 megawatts are set to come online fully in 2013, and SunPower’s 250 megawatts should be running by 2012, both in central San Luis Obispo County. Unlike these photovoltaic projects, most large-scale solar farms feature solar thermal systems.

I found a story in the Milwaukee Business Journal that mentioned some of the area requirements:

The 550-megawatt Topaz Solar Farm to be built by OptiSolar, a venture-backed firm based in Hayward, will utilize OptiSolar’s relatively low-cost thin-film photovoltaic panels manufactured in Hayward and Sacramento. The project, which will cover about 9 square miles with solar panels, will be the largest solar PV project in the world, and is reported to cost $1 billion. It is expected to begin power delivery in 2011 and be fully operational in 2013.

How does that area requirement compare with the area I came up with in A Solar Thought Experiment for providing electricity for the entire U.S.? 550 megawatts over 9 square miles is equivalent to 62 megawatts per square mile. My calculations were that it would take 2531 square miles (a square of about 50 miles by 50 miles) to provide 882,125 megawatts (349 megawatts per square mile). If we scale that down to 9 square miles, we would come up with (882,125 megawatts/2531 square miles) * 9 square miles = 3,100 megawatts (versus their reported 550 megawatts). I would then suspect that they are taking the infrastructure, etc. into account, as my calculation was purely surface area of the panels.

However, I did a second calculation in Running the U.S. on Solar Power in which I did consider the total land requirements of a solar thermal plant. Apples and oranges, I know, but I want to know how they stack up against each other. For the Optisolar plant, we find 550 megawatts on 9 square miles. A square mile contains 640 acres, thus we have 0.89 megawatts per acre. In my previous calculation on solar thermal, I found (based on an actual plant capacity and land requirements) that they expected to get 0.147 megawatts per acre.

That suggests one of three things to me. First, it could mean that the supporting infrastructure for a solar thermal plant has a much greater requirement than a solar PV plant. Second, it could mean that I am looking at just PV surface area versus total infrastructure requirements for solar thermal, but that doesn’t seem to mesh with my prior calculation (unless they are usually very low efficiency panels in the PG&E projects). Or, it could be that since I am only looking at peak output, I am not actually getting a total picture since solar thermal has the potential for energy storage, and thus generating power far past peak output. In other words, solar PV has a pretty sharp peak, and solar PV has a much broader peak for the generation of electricity. There is a 4th option – that I have an error somewhere. But I know if I throw this out there, someone will sniff it out.

Maybe some of our knowledgeable solar experts can weigh in?

Finally, the inevitable question always seems to be whether you can invest in the companies involved. According to this article, Optisolar has been raising capital through private investors. SunPower, however, trades on the NASDAQ as SPWR. The PE is currently at 142, so if you are risk averse you are probably going to pass. The other company mentioned in the article, Pacific Gas and Electric, trades on the NYSE as PCG and a more conservative PE of 14.8.

August 15, 2008 Posted by | investing, solar PV, solar thermal | 192 Comments

Replacing Gasoline with Solar Power

Executive Summary

If you don’t want to run through the calculations, here is the summary. I attempted a thought experiment in which I calculated whether it would be feasible to use solar power to generate enough energy to offset all U.S. gasoline consumption. My conclusion is that it will take about 444,000 megawatts of electrical generating capacity. Current U.S. generating capacity is over 900,000 megawatts, but there isn’t a whole lot of spare capacity in that number.

To generate 444,000 megawatts with solar PV would require just under 1,300 square miles (a 36 mile by 36 mile square) of just PV surface area. To generate that much power with solar thermal – including supporting infrastructure – would require 4,719 square miles (a 69 mile by 69 mile square). A large area, but not impossible to envision us eventually achieving this.



Having made an attempt to calculate the number of square miles to replace current U.S. electricity consumption via solar PV or solar thermal, I have been challenged to do the same exercise for replacing our gasoline usage. (In fact, I was told by someone that they had never seen this kind of calculation done, so I told them I would do it). I have no idea how this calculation is going to turn out, but I suspect it is going to be similar to the previous calculation for replacing electrical consumption. My guess is less than 100 miles by 100 miles. Note that this is a thought experiment, in which I try to get an idea of what it would take to achieve this.

First, some caveats. There are still technical obstacles that prevent this scenario from being realized. Those are, 1). Battery range is still too low (The plug-in Prius is only going to be able to go 7 miles on battery power).; and 2). Solar power can’t be adequately stored. However, that’s not the purpose of the exercise. The purpose is to satisfy my curiosity: If we were going to try to replace gasoline with solar power, are the land requirements prohibitive?

I am only going to do this calculation for gasoline, as I think it is unlikely that electricity will ever power long-haul trucks or airplanes.

How Much Do We Need?

The U.S. currently consumes 389 million gallons of gasoline per day. (Source: EIA). A gallon of gasoline contains about 115,000 BTUs. (Source: EPA). The energy content of this is equivalent to 45 trillion BTUs per day. The average efficiency of an internal combustion engine (ICE) – that is the percentage of those BTUs that actually go into moving the vehicle down the road – is about 15%. (Source: DOE). Therefore, the energy that goes toward actually moving the vehicles is 6.7 trillion BTUs per day.

The efficiency of electric infrastructure can be broken down into several steps. According to this source, the respective efficiencies for the transmission lines, charging, and the vehicle efficiency are 95%, 88%, and 88%, for an overall efficiency (after the electricity is produced) of 74%. To replace the gasoline BTUs that go toward moving the vehicle with electricity is going to require 6.7 trillion/0.74, or 9.1 trillion BTUs. To convert to electricity, we use 3,413 BTUs/kilowatt-hour (kWh). Thus, 9.1 trillion BTUs/day is equal to 2.7 billion kWh/day. That’s how much energy we need. To convert this to power, we need to multiply by 1 day/24 hours, and that gives us 111 million kilowatts, or 111,000 megawatts (MW) of power generation required.

Looking back at my Solar Thought Experiment, I calculated 2,531 square miles to replace our peak electrical demand of 746,470 MW (746 GW). However, the current calculation is a different sort of calculation than what I did previously. The previous calculation attempted to have enough installed solar PV to meet peak demand. In the case of replacing our transportation fuel, I need enough panels to produce the required transportation energy in 8 hours or so while the sun is shining. To be conservative, we can assume 6 hours, which means we will actually need four times the 111,000 MW, or 444,000 MW.

Using Solar PV

From the previous essay, I used a conservative value of 12.5 watts per square foot as the generating capacity of an actual GE PV panel. To get 444,000 MW is going to take an area of 35.5 billion square feet, which is 1274 square miles. This is an area of just under 36 miles by 36 miles. However, this is just the surface area required to generate the electricity. It does not include area required for supporting infrastructure.

Using Solar Thermal

Doing the same calculation based on the solar thermal output from Running the U.S. on Solar Power, the expectation was that 0.147 megawatts could be produced per acre. This did include all of the land associated with infrastructure. If we use that number, we find that to generate 444,000 MW is going to take a little over 3 million acres, or 4,719 square miles. This is a square of just under 69 miles by 69 miles.

The reality is that we would use a combination of the solar PV and solar thermal. We have a lot of available rooftops that can create electricity with solar PV, and there are large tracts of land in sunny Arizona and Nevada that can create electricity with solar thermal.


Clearly, a lot of area is required, but it isn’t impossibly large. Of course to achieve this, a couple of big problems need to be resolved. First, battery life needs to improve somewhat before people are going to embrace electric transport. According to this ABC News story, the average commute is 16 miles one-way, but the range of the plug-in Prius is only expected to be 7 miles. The Aptera, on the other hand, claims a range of 120 miles. Maybe we just need to change the way we think about what we drive. (On the other hand, not a lot of commuters are going to climb into an Aptera if they have to share the road with large SUVs).

Second, and the bigger issue, is that we still don’t have a good way to store excess solar power. We need to have a good storage mechanism so electric cars can be charged at night from solar electricity produced during the day. One idea for this that I have seen floated is to use peak solar energy to electrolyze water, and then store the hydrogen in centralized locations. The hydrogen would then be burned at night to run centralized electrical generators. Not the most efficient method for storing solar energy, but technically workable.

Finally, the current electrical grid couldn’t handle such a large increase, but the model I envision would generate and consume the electricity locally.


I had delayed posting this for almost a week, because I was sure there was an error in the calculations. I finally found one (I had turned a kilowatt into a watt), but let me know if you find other errors or incorrect assumptions.

May 12, 2008 Posted by | electric cars, electricity, electricity usage, solar power, solar PV, solar thermal | 107 Comments

How to Change the World

Fortune has a very interesting interview with Google co-founder Larry Page. He hits on a lot of topics that are frequently discussed here, and some that aren’t often discussed, but that I have spent a lot of time thinking about (e.g., geothermal). Here is a link to the interview:

Larry Page on how to change the world

And some energy-specific excerpts:

Do you have other examples where innovative leadership could move the needle?

I think there are a lot of areas. You can be a bit of a detective and ask, What are the industries where things haven’t changed much in 50 years? We’ve been looking a little at geothermal power. And you start thinking about it, and you say, Well, a couple of miles under this spot or almost any other place in the world, it’s pretty darn hot. How hard should it be to dig a really deep hole? We’ve been drilling for a long time, mostly for oil – and oil’s expensive. If you want to move heat around, you need bigger holes. The technology just hasn’t been developed for extracting heat. I imagine there’s pretty good odds that’s possible.

Solar thermal’s another area we’ve been working on; the numbers there are just astounding. In Southern California or Nevada, on a day with an average amount of sun, you can generate 800 megawatts on one square mile. And 800 megawatts is actually a lot. A nuclear plant is about 2,000 megawatts.

The amount of land that’s required to power the entire U.S. with electricity is something like 100 miles by 100 miles [RR comment: That’s around what I have come up with whenever I tried to calculate it. Maybe that’s where he got it, since I often get hits from Google in Mountain View. 🙂] So you say, “What do I need to do to generate that power?” You could buy solar cells. The problem is, at today’s solar prices you’d need trillions of dollars to generate all the electricity in the U.S. Then you say, “Well, how much do mirrors cost?” And it turns out you can buy pieces of glass and a mirror and you can cover those areas for not that much money. Somehow the world is not doing a good job of making this stuff available. As a society, on the larger questions we have, we’re not making reasonable progress.

And it looks like we are on the same page – no pun intended – regarding the solution to our energy problems:

So you think that geothermal and solar thermal could solve our energy problems?

Yeah, probably either one could generate all the energy we need. There’s no discipline to actually do this stuff, and you can also see this vested interest, risk-averse behavior, plus a lack of creativity. It sort of conspires. It’s also a timeliness thing; everyone said Sam Walton was crazy to build big stores in small towns. Almost everyone who has had an idea that’s somewhat revolutionary or wildly successful was first told they’re insane.

He also comments on who needs to be working on these changes:

Whose obligation is it to make this kind of change happen? Is it Google’s? The government’s? Stanford’s? Kleiner Perkins’?

I think it’s everybody who cares about making progress in the world. Let’s say there are 10,000 people working on these things. If we make that 100,000, we’ll probably get 10 times the progress.

And then you compare it with the number of engineers at Exxon and Chevron and ConocoPhillips who are trying to squeeze the last drop of oil out of somewhere, and all the science brainpower that’s going to that. It’s totally disproportionate to the return that they could get elsewhere.

What kind of background do you think is required to push these kinds of changes?

I think you need an engineering education where you can evaluate the alternatives. For example, are fuel cells a reasonable way to go or not? For that, you need a pretty general engineering and scientific education, which is not traditionally what happens. That’s not how I was trained. I was trained as a computer engineer. So I understand how to build computers, how to make software. I’ve learned on my own a lot of other things. If you look at the people who have high impact, they have pretty general knowledge. They don’t have a really narrowly focused education.

April 30, 2008 Posted by | geothermal, Google, Larry Page, solar power, solar thermal | 13 Comments