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Will Solar Prices Fall into Grid Parity?

The following is a guest post written by Dan Harding. Dan has written numerous articles on the solar industry, and is a regular contributing author to CalFinder.
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Will Solar Prices Fall into Grid Parity? 


By Dan Harding

The Holy Grail…in solar-speak, it translates roughly to Grid Parity. It is a goal either mythical or predestined, depending on which side of the solar power movement the speaker resides. A recent surge in supply and technology, coupled with increased government subsidies, are tipping the scales toward destiny, although by no means is the path to grid parity set in stone. The rapid fall in prices for solar panels and other system components in an oversupplied and flooded market could continue home solar power on its way to that mythical Grail but, all mythos and wishful thinking aside, what are the odds?

Good, says Swami Venkataraman, Director of Corporate and Government Ratings at Standard & Poor’s, in a recent assessment of the U.S. solar market for Renewable Energy World. As of February, 2009, installed costs for residential and commercial photovoltaic (PV) systems had fallen to $7.60 per watt from $10.50 per watt just two years earlier. Prices continued to fall throughout 2009 and, while expected to stabilize somewhat as the national economy rebounds, they should remain on that downward slope in 2010 and beyond.

So when will solar cross that line? It could be soon, very soon in regions of the country with either abundant sunlight (southwest) or relatively high electricity costs (northeast). Yet some valuable help is still needed at the legislative level which, if provided, could propel solar power to grid parity in the short-term in the aforementioned regions.  


Three factors, says Venkataraman, can help make PV cheaper than, say, a combined-cycle gas turbine plant. One or all of the following could ensure solar power a level playing field in the long term:
  • Rising gas prices
  • Renewable portfolio standards that make renewable energy credits (RECs) more valuable
  • The passage of carbon legislation that would force gas power producers to buy carbon credits, thus forcing an increase in price for natural gas.

Including incentives, solar power is already close to grid parity in many areas. The Northeast holds the handy combination of some of the most lucrative solar incentives (per watt installed) in the country, as well as the highest electricity prices. Therefore, solar has far less distance to make up to reach at least natural gas, and gives solar power the best and fastest chance to reach grid parity in the nation. In California, where incentives have been declining for several years now, the primary advantage is in abundant sunlight (same goes for Arizona, New Mexico, west Texas, etc.), as well as a powerful RPS and a general eagerness from the public to adopt clean energy.

But as those two examples illustrate, grid parity will almost certainly NOT come to the United States as a whole all at once. Federal incentives were expanded in 2009, including the removal of the $2,000 cap on residential systems and the admittance of utilities into the Investment Tax Credit, but continue to vary widely between states. The feds provide a baseline subsidy, but what truly makes solar affordable for most homeowners and businesses are the added incentives offered by their state. So, in terms of reaching grid parity, we can expect the Southeast — despite its healthy share of sunshine — to be the slowest to reach the Holy Grail. This is due primarily to a lack of incentives, low electricity costs and a deep connection to fossil-fueled electricity.

Without incentives, there is still a real chance for PV, especially commercial PV, to reach grid parity in the relative short-term. Current capital costs for commercial PV are about $5.50 to $6.60 per watt depending on the size of the installation, according to Standard & Poor’s. Incentive levels in many northeastern states are upwards of $4.00 per watt, which means that, given incentives, the levelized cost of electricity (LCOE) of commercial PV systems was already below standard commercial rates. Furthermore, if falling panel prices enable systems to reach or fall below $5.00 per watt, then solar PV could reach parity even without subsidies.

Residential grid parity is more distant but still closest in the Northeast. Outside of the Southwest and Northeast, where solar irradiance and/or electricity costs make the solar-grid-parity question more complicated and uncertain, help will have to come from other renewables. Most notable among these are geothermal (Northwest) and wind power (Midwest). It is important when discussing grid parity for solar power not to forget its intermittency and the fact that some backup power system will be needed. Even if our solar infrastructure were so advanced as to provide all our power needs during peak load times, we would still need alternative sources to pick up the slack on cloudy days and at night.

Of course, straight-laced economics aside, we must also consider the inherent value of solar power beyond mere dollar signs. The point of renewable energy is to switch from pollutive, peaking sources of energy to clean, renewable ones. Solar power emits no greenhouse gases, no carbon dioxide and, when distributed, can provide power at or near the point of use without turning our cities into smog factories. That alone is reason enough to subsidize solar, wind, geothermal and other renewable resources until they reach the Holy Grail that is their destiny.

March 3, 2010 Posted by | guest post, reader submission, solar power, solar PV | 14 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.

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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!

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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.

http://rredc.nrel.gov/solar/old_data/nsrdb/1961-1990/redbook/sum2/

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.

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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

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

Largest Affordable Solar Housing Installation

Most press releases I receive are not really worth publicizing, in my opinion. Most of them are just self-promotions for one organization or another. But here’s one that is worth bringing some attention to:

The Largest, Solar-Powered, Affordable Housing Community in the U.S.

24 Buildings with 378 Family Apartments at Crescent Park in Richmond, California

Dedication: Tuesday, June 10: 10 AM – 12 NOON

The Crescent Park Apartments, 5000 Hartnett Avenue, Richmond, CA

Richmond, California (June 10, 2008) – The new solar installation at the Crescent Park apartments in Richmond, California will be dedicated on Tuesday, June 10, making this large apartment complex the largest, solar-powered affordable housing community in the United States.

EAH Housing, a nonprofit affordable housing developer and manager based in San Rafael California, has announced the dedication of the solar system installed by Sun Light and Power serving 24 buildings and 378 family apartments for this large community on nearly 25 acres.

About the Solar Energy Installation

The solar installation at Crescent Park – a $7 million dollar project installed by Sun Light and Power, based in Berkeley, CA – will help reduce the production of greenhouse gases while providing lower utility costs for this large, family complex. The installation also helps the city of Richmond to meet almost 20% of its 5 MW goal for usage of solar power.

Producing nearly a megawatt of clean, renewable energy, the 908 kilowatts system includes 4323 SunPower, 210-watt modules and 180, SunPower M Series inverters.

Gary Gerber, President of Berkeley’s Sun Light and Power said: “I am honored and pleased that EAH chose Sun Light & Power to build America’s largest affordable housing solar project here at Crescent Park. Sun Light & Power’s engineering expertise and our 32 years of experience designing and installing solar systems made us a great match with EAH, who clearly recognize that solar electricity must play a pivotal role in providing the renewable energy this country needs to reduce our environmental impact and stabilize our future energy costs.”

About the Crescent Park Apartments

Crescent Park is currently undergoing a $70 million restoration by EAH Housing.

EAH Housing acquired Crescent Park in 1994, and quickly moved to restore it to its former status as a valued community resource. It has since been awarded the HUD “Best Practices” Award for its Computer Learning Center, which provides access to technology to very low-income residents.

“We are proud to provide solar power for all 378 apartments as part of our major rehabilitation and improvement of this important affordable housing community that has served Richmond since the 1960s,” said Mary Murtagh, President and CEO of EAH Housing.

“The contributions made by EAH Housing to the city of Richmond is incalculable,” said Gayle McLaughlin, Mayor of Richmond. “EAH has created an affordable housing complex that not only brings value to our city but, with its commitment to solar energy, has also taken on a stewardship role as well. We look forward to our continuing relationship with this outstanding organization.”

About Sun Light and Power

Sun Light and Power was formed 32 years ago in Berkeley by Gary Gerber as one of the first solar energy companies in California. Now the 12th largest solar aggregator in the U.S. (with more than 600 solar installations in the Bay Area) and a staff of 54 employees, Sun Light and Power prides itself on its skilled engineering staff who work with many technically-challenging projects. More information is available at: http://www.sunlightandpower.com/ or by calling: (510) 845-2997.

About EAH Housing

EAH Housing is one of the most respected nonprofit developers/managers in the western United States. With properties in 14 counties in California and Hawaii, and over 40 years of providing affordable housing, EAH Housing has developed over 5900 homes and manages 75 properties. EAH serves over thousands of families, seniors, students and persons with disabilities in California and Hawaii. EAH has regional offices in San Rafael, Fresno, San Jose, and Honolulu. More information is available at: http://www.eahhousing.org/.

NEWS NOTE: Gary Gerber, founder and president of Sun Light and Power (Berkeley, CA) is available to talk with you. His contact information: Phone: (510) 845-2997. Gary@sunlightandpower.com

June 7, 2008 Posted by | solar power, solar PV | 11 Comments

Lumeta: Peel and Stick Solar

I wanted to thank a reader for sending me the following YouTube link. This is very cool. A solar company called Lumeta – a subsidiary of California-based DRI Companies, has created a peel and stick solar panel. Watch just a bit of this video, as two men install 2.25 kilowatts in 34 minutes. Makes it look easy.

That could produce more than enough power for the average daily commute, but I couldn’t find any details on costs nor of panel efficiencies.

I am putting this up in a hurry, as I am in London today and tomorrow to talk about green technologies with a couple of groups. It will be later in the week before I have a chance to put anything else up.

May 19, 2008 Posted by | solar power, solar PV | 21 Comments

Lumeta: Peel and Stick Solar

I wanted to thank a reader for sending me the following YouTube link. This is very cool. A solar company called Lumeta – a subsidiary of California-based DRI Companies, has created a peel and stick solar panel. Watch just a bit of this video, as two men install 2.25 kilowatts in 34 minutes. Makes it look easy.

That could produce more than enough power for the average daily commute, but I couldn’t find any details on costs nor of panel efficiencies.

I am putting this up in a hurry, as I am in London today and tomorrow to talk about green technologies with a couple of groups. It will be later in the week before I have a chance to put anything else up.

May 19, 2008 Posted by | solar power, solar PV | 21 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.

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Introduction

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.

Conclusions

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.

Note

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