Posts Tagged ‘Solar energy’


Comparison of Arizona Nuclear and Solar Energy

December 9, 2015

Let’s compare and contrast solar energy and nuclear energy in Arizona. There is only one nuclear power plant in the state, the Palo Verde Nuclear Generating Station in Tonopah. There are several solar energy sites, so we will pick the Aqua Caliente Solar Project because it won the Renewable Energy World Solar Project of the Year category in their 2012 Excellence in Renewable Energy Awards.

Palo Verde Nuclear Generating Station

This nuclear plant consists of three reactors with with a total nameplate capacity of 3,937 MW. If these reactors ran for 24 hours day for 365 days a year they would yield 34,500 GWh (gigawatt hours) per year. The actual output is about 31,300 GWh per year (2010). This means they have a capacity factor of about 90%. Averaged over time Palo Verde yields 3,543 MW.

Palo Verde became operational in 1988 and is currently approved to operate until 2047, giving a lifetime of nearly 60 years.

Palo Verde’s construction cost was $5.9 billion in 1988 ($11.86 billion in 2015 dollars). Its operating costs for fuel and maintenance were about 1.33 cents per kWh in 2004 (1.67 cents in 2015 dollars.)

Based on an average power yield of 3,543 W and a cost of $11.86 billion (in 2015 dollars), the construction cost per watt for Palo Verde was $3.34 per Watt (in 2015 dollars).

Agua Caliente Solar Project

This 9.7 square kilometer solar energy farm has a nameplate capacity of 290 MW peak.  Its first year of full operation was 2014. If it were able to produce its nameplate capacity of 290 MW continuously for one year the energy output would be 2540 GWh. The energy output was 741 GWh in 2014, which means a capacity factor of 29%, an excellent result for solar energy. Averaged over time, this solar farm yields 84.6 MW.

Construction cost for Aqua Caliente was $1.8 billion.

Based on an average yield of 84 MW and a construction cost of $1.8 billion, the construction cost per watt for Aqua Caliente was $21.43 per Watt.


The cost per kilowatt hour of energy for either of these sources is combination of the construction cost and the operation, fuel and maintenance cost.  The longer the facilities are in operation the lower the fraction of construction cost per kilowatt hour.

The operation, fuel and maintenance cost for the Palo Verde Nuclear plant were about 1.33 cents per kWh in 2004 (1.67 cents in 2015 dollars.)  The great advantage of the Agua Caliente solar farm is that its fuel cost is zero, and we will assume for the sake of argument that its other operation and maintenance costs are also zero.

The following chart shows various costs per kilowatt hour for each of the facilities for various lifetimes.


1.  $0.0133 per kilowatt hour in 2004.  Converted to 2015 dollars.
2. 2013 energy output.
3. $5.9 million construction cost in 1988 dollars.  Converted to 2015 dollars.
4. 2014 energy output
5. $1.8 billion construction cost in 2014.
6. (GWh/year) x (number of years) x (1,000,000)
7. (Construction cost) / (kilowatt hours produced over lifetime)
8. (Construction cost per kWh) + (operating cost per kWh)

Two blocks of data are highlighted in yellow.  These are the most likely lifetime scenarios for each of the power generating plants.  The Palo Verde nuclear plant has had its license extended to 60 years.  Aqua Caliente solar farm is made from First Solar CdTe modules that have a 10 year material and workmanship warranty and a  warranty of 80% of the nominal output power rating during twenty-five (25) years.  It is reasonable to hope that it will last 40 years

There is one more thing to be considered.  We have assumed so far that the yearly output of each of these power generating stations it the same year after year.  That is not entirely correct.  Historically, the Palo Verde nuclear plant has increased its capacity factor through time as operations have become more efficient.  Whether that trend will continue is unknown.

Solar modules tend to slowly degrade with time.  The First Solar CdTe modules that are used at Aqua Caliente will likely decay at about 0.5% per year. The chart above gives a best case estimate for Agua Caliente and does not compensate for this degradation.

Based on the highlighted sections of the above chart, Aqua Caliente Solar Farm will likely cost about 2.5 times more per kilowatt hour than the Palo Verde Nuclear Plant over the course of their lifetimes.

One more point.  Aqua Caliente requires 9.7 square kilometers to generate an average of 84.6 MW.  Palo Verde Nuclear Plant generates and average of 3,543 MW.  So it would take 41 Agua Calientes to equal the power of Palo Verde.  That would require about 400 square kilometers.

Energy is the lifeblood of civilization.  The pursuit of energy abundance is the pursuit of healthier and more fulfilling lifestyle for greater numbers of people.  I present this data to help inform the choices that need to be made in that pursuit.


How much photovoltaics to provide 100 kilowatt hours per person per day?

November 8, 2015

Suppose you wanted to power the world at the level that each human being can enjoy the same level of energy abundance as the average American. And suppose we wanted to do it all with photovoltaic solar energy. What would it take?

There are an average of 250 kilowatt hours consumed per person per day in the United states. Maybe that seems like a lot to you because you occasionally look at your home electric bill and see less than 1000 kilowatt hours used in a entire month for a home that houses four people. That 1000 kilowatt hours for four people in a month works out to only about eight kilowatt hours per person per day. But that electric bill is a very poor indicator of how much energy is actually expended for your benefit. That is why claims that some energy source will power X number of homes is incredibly misleading.

Here is the reality.  According to Lawrence Livermore National Laboratory the United States consumes 98.3 quads of energy every year.


That works out to about 250 kilowatt hours per person per day

(98.3 quads/year) x (2.933 x 1011 kw-hr/quad) / (year / 365 days) / (3.2 x 106 people) = 247 kilowatt hours/person/day

Fortunately, this daunting amount of energy is also somewhat misleading.  Look at the right side of the graph from Lawrence Livermore.  Notice that the two final energy outputs on the right side of the graph are “Energy Services” and “Rejected Energy.”  “Energy Services” is energy that actually does some useful work.  “Rejected Energy” is energy that is lost, mostly in form of waste heat.  For example, if you burn a lump of coal in a steam generator and get a kilowatt of energy out in the form of electricity, but lose two kilowatts in the form of heat to the atmosphere, then you got one kilowatt hour of Energy Service but two kilowatt hours of Rejected Energy.  As you can see from the graph, only 40% of the energy that is input comes out the the system as Energy Services (38.9 quads / 98.3 quads).

One of the big advantages of solar photovoltaics is that you don’t lose 60% of your energy to heat.  Electric cars put far more of their stored electric energy into useful work (Energy Services) and far less into “Rejected Energy” than do blazing hot internal combustion engines.

Let’s make the assumption for now that every possible efficiency is applied, so that we only need to produce 40% of the 250 kilowatt hours per day per person, or 100 kilowatt hours per day per person.  Still a lot of energy, but more manageable than 250 kilowatt hours.

So, for 7 billion people we need 700 billion kilowatt hours per day (100 kilowatt hours per person x 7 billion people).  If we got all that energy from solar photovoltiacs, how much land would be required for solar arrays, how much would it cost?

Topaz Solar Farm

To get estimates of these values, we can look at some of the world’s biggest solar arrays.  Consider the Topaz Solar Farm in California.  It is one of the biggest and one of the newest in the world and in an area of very high solar insolation.  It is expected to generate 1,100 GWh of energy per year while occupying 25 km2 with a cost of $2.5 billion.  Therefore it would generate the energy consumed by about 30,000 people at 100 kWh per person per day.

(1100 GWh/year)x(1×106 kWh/GWh)x(year/365 days)/(100 kWh/person/day) = 30,136 people

From this it is clear that it would take about 6 million km2 of solar photovoltaics of the Topaz Solar Farm density to generate all the energy consumed by 7 billion adequately powered people.

(7×109 people) / (30,136 people/25 km2) = 5.8×106 km2

Keeping in mind that the Topaz Solar Farm cost $2.5 billion and yields enough energy for 30,136 people, then the cost for 7 billion people would be about $580 trillion.

(7×109 people) / (30,136 people/$2.5×109) = $5.8×1014 .

For the sake of comparison the, the gross domestic product of the United States is about $17 trillion, or less that 3% of that $580 trillion.  The gross product of the entire world  is about $78 trillion, or about 13% of that $580 trillion.  So, if every penny or mark or yen, etc. of world product for about 7.5 years were dedicated to this project, it could be accomplished.

Some points to consider

What would be the consequences of covering 6 million square kilometers of land with PV?  This would be like completely covering an area the combined size of Arizona, Nevada, Colorado, Wyoming, Oregon, Idaho, Utah, Kansas, Minnesota, Nebraska, South Dakota, North Dakota, Missouri, Oklahoma, Washington, Georgia, Michigan, Iowa, Illinois, Wisconsin, Florida, Arkansas, Alabama, North Carolina, New York, Mississippi, Pennsylvania, Louisiana, Tennessee, Ohio, Virginia, Kentucky, Indiana, Maine, South Carolina, West Virginia, Maryland, Vermont, New Hampshire, Massachusetts, New Jersey, Hawaii, Connecticut, Puerto Rico, Delaware, Rhode Island with solar panels.  Of course, this would be spread out over the about 100 million square kilometers of land at latitudes lower than about 50 degrees.

This plan would also require a distribution system that could move energy from daytime areas to nighttime areas, or at least a few days of storage for every person on the planet.  Such a distribution system is not feasible at this time, and the massive amount of storage is prohibitively expensive.

Two days of storage would be 200 kilowatt hours of stored energy per person.  Probably the best mass storage option today (2015) is with Tesla’s Powerwall, which stores 7 kilowatt hours, costs $3,000, and weights 220 pounds.  So we would need about $90,000 and about 6,600 pounds of storage for each of the 7 billion people.  That adds another $630 trillion to the cost.

These calculations serve simply to give a feel for what could be done with solar photovoltaics and what the limitations might be.  I am not suggesting that the world should be powered solely with PV.  With other energy sources in the mix less money and land would need to be devoted to PV (but more to those other sources).  For example, if you did the same calculations for wind, then you would find that about twice as much area  (about 12 million square kilometers) would have to be covered by wind farms to get the same amount of energy.  But at least you can grow corn are graze cattle below the turbines in a wind farm.

I have led you to water.  It is up to you to drink up your own conclusions about the viability of using solar energy to bring the world up to a reasonable level of energy consumption.


News: White house installs 6.3 kW solar PV system

May 28, 2014

You’ve probably seen the news – it been reported everywhere: the White House has installed 6.3 kilowatts of solar PV.  How much energy will that system yield in practical terms?

The rated power of a PV system is the power that it will yield when the sun shines straight on to it with an irradiance of 1000 W/m2, and the temperature is 25 degrees C .  If those conditions are met for one hour, then the white house PV system will yield 6.3 kilowatt-hours.  If they are met for two hours, then it will yield 12.6 kilowatt-hours, etc.

But those conditions are rarely met exactly.  In the United States those conditions will  be approximately met around noon time on a cold sunny day if the panels are mounted at the proper angle.  Your results will vary.

There is another way to approximate the average daily energy that solar panels will yield: multiply the rated value of the panels by the average daily insolation.   The accuracy of this approach depends on the angle that the panels are mounted and several other variables, but this very simple approach will give an approximation good enough for our purposes

The average daily insolation in Washington DC is 4.23 kilowatt-hours/m2/day.  So the White House PV system would yield about 27 kilowatt-hours per day.

How much energy is 27 kilowatt-hours in practical terms?

If you add up all he energy that is consumed in the United States each year and divide it by the population and 365 days, then you will get the average daily per capita energy consumption in the United States.  Many people have a vague (and wrong) idea of what their energy consumption is, perhaps based on their monthly electricity bill.  They are often shocked when they learn the truth.

Here are the numbers…

Total US yearly energy consumption: 97.4 Quads

This is equal to 28,560,000,000,000 kilowatt-hours (since one Quad is 293,000,000,000 kilowatt-hours).

This translates to a daily per capita consumption is 247 kilowatt-hours per day


Therefore, the solar PV on the roof of the White House provides about 10% of the 247 kilowatt-hours consumed by the average American each day!

In all fairness though, we have to understand that a large portion of that 247 kilowatt-hours per day per capita does not do any useful work.  I hesitate to say that this large portion is “wasted.”  Rather, it is lost mostly due to the inescapable consequences of thermodynamics.  Look at the energy flow chart from Lawrence Livermore National Laboratory.  On the right side you can see that out of the 97.4 Quads of energy used yearly in the United States, 38.4 Quads goes to “energy services” and 59.0 Quads are “rejected energy.”  Effectively, 97.4 Quads of energy are consumed in order to yield 38.4 Quads of “energy services.”  That is only about 40% efficient.

Almost all of the energy generated by the Solar PV system on the roof of the White House can be used for “energy services.”  With this in mind we could fairly claim that the PV system on the roof of the White House provides about 30% of the energy needs (or “energy services”) of a single average American.

Some more perspective on 27 kilowatt-hours of energy

A gallon of gasoline has an energy content of about 32 kilowatt-hours.  If you drive a truck that gets 15 miles per gallon, then you are consuming about 2.1 kilowatt hours per mile [ (32 kilowatt-hours / gallon) / (15 miles / gallon) ].  If you are driving down the highway at 60 miles per hour, then you will consume 27 kilowatt-hours in a mere 13 minutes!

If you are driving down the road at 60 mph in the President’s limo, which gets only 8 mpg, then you will burn up your 27 kilowatt hours in only 7 minutes.  Of course, the President travels with an entourage of about 45 vehicles along with his limo.  It is a pretty good bet that most of these vehicles are heavy-duty, low mileage vehicles.  I think a good approximation would be an average of 20 miles per gallon.  At 60 miles per hour this entourage would burn about 135 gallons an hour, or about 4320 kilo-watt hours per hour.  At that rate they would burn the allotted 27 kilowatt-hours every 22.5 seconds

Consider the Boeing 747-200B (or its militarized version: the VC25, such as Air Force One).  It has a range of 6,100 miles on 48,445 gallons of fuel and a typical cruising speed of 555 mph.  That works out to about 8 gallons per mile (that is “gallons per mile” not “miles per gallon!”) and about 9 miles per minute, resulting in a fuel consumption rate of about 72 gallons per minute!  The energy content of the jet fuel is about the same as gasoline, about 32 kilowatt-hours per gallon.  So, in one minute of cruising the 747 consumes about 2300 kilowatt-hours of energy.


At that rate, the 747 will consume a days worth of the energy produced by the White House solar PV system in about 0.7 seconds (after traveling only about 500 feet), and a years worth of energy in about 4 minutes and 15 seconds.  A round trip from, say, Washington DC to Hawaii and back is about 9540 miles.  At 550 mile per hour that would be about 17.3 hours of flight.  How long would the White House Solar PV array have to operate to produce enough energy for that round trip time?  Answer: 243 years.

243 years

My position

I have been working on solar PV research for about 17 years and believe any energy source is worthy of research.  But religious devotion to one source or another does not advance the human race.  Consequently, I am a supporter of both nuclear energy and solar energy (if and when it is affordable and competitive on its own merits – not massive subsidies).


Units of energy: homes?

March 8, 2014

corrected 4/12/14

How many BTUs are in a kilowatt-hour?  How many barrels of oil equivalent (BOE) are in a kiloton of TNT?  There are a lot of different units of energy and power.  Which one is chosen at a particular time depends on the field and the customs of its experts.  It can get a little confusing when comparing numbers from practitioners in different fields.

It can be very eye opening to make the conversions.  For example, six sixteen watt CFL bulbs lit up for six hours will use as much energy as released by the detonation of one pound of TNT.  My preference is to convert powers to watts and  energies to watt-hours.

New unit for power

But there seems to be a new unit of power that I can’t find in any of my physics books.  Its called a “home.”  Here are some examples of its usage…

“The Tatanka Wind Farm, on the North Dakota-South Dakota border, will power 60,000 homes.”

“Limon I Wind Energy Center in Colorado is capable of generating enough electricity to power approximately 100,000 homes.”

“[E]nough clean electricity to power over 60,000 homes.”

“A 230 MW photovoltaic solar station in the Antelope Valley of California that will supply enough energy for 70,000 homes.”

“The new Copper Mountain 3 solar plant, which will be finished in 2015, will be able to generate enough power to supply around 80,000 homes.”

“Chicken Manure to power 90,000 Homes in the Netherlands!”


Ivanpah mirrors

Mirrors at Ivanpah

Brightsource’s Ivanpah Solar Electric Generating System in California is a case in point.  This is a solar thermal site that uses thousands of mirrors to concentrate sunlight to generate heat to run generators. says  the “$2.2 billion Ivanpah Solar Electric Generating System—the largest of its type in the world—will power 140,000 California homes.”  It looks like they are using a “home” as a unit of power.

What does “will power 140,000 California homes” really mean?

According to the EIA, the average home in California consumes about 7000 kilowatt-hours of electric energy each year  (most recent data, 2009).  That means 140,000 homes would use 9.8 x 108 kilowatt-hours (9.8 x 105 megawatt-hours) of electric energy per year.  I think we’re on the right track here, because the National Renewable Energy Laboratory says Ivanpah will produce 10.8 x 105 megawatt-hours per year.

But this unit of power called a “home”  is still a little misleading.  Although the average California home consumes about 7000 kilowatt-hours of electric energy per year, energy from other sources is also consumed.  The other big source is natural gas, which may be used for space heating, cooking or water heating.  If you think this is trivial compared to the amount of electricity used, think again.  The EIA document on residential energy consumption in California shows these graphs…

EIA California energy consumption

I think it is bad practice to use two mix different units for energy (kilowatt-hours and Btu) as the EIA has done with these graphs.  How many people can compare kilowatt-hours and Btu by looking a graphs?

The graph on the top left is where I got the estimate of 7000 kilowatt-hours of electrical energy per year for the average California home.  Notice that it is labled “ELECTRICITY ONLY.”  The graph on the lower left is for “ALL ENERGY average per household,” and indicates about 62 million Btu per California home per year.

How does 62 million Btu compare to 7000 kilowatt-hours?   62 million Btu translates to 18,170 kilowatt-hours!  In other words, 11,170 kilowatt-hours of energy consumed in the average California home comes from sources other than electricity.  If you find this hard to believe, look at the number of kilowatt-hours you used on a recent winter electric bill and look at the amount of energy, usually in “therms,” on a recent winter gas bill.  Convert the “therms” to kilowatt-hours and you will see what I mean.  It takes a lot more energy to heat water and air in your house than it does to light your bulbs or power your TV.  So Ivanpah really only provides enough energy to power 54,000 (≈140,000 x (7000/18,170)) California “homes.”

You might think that providing enough energy for 54,000 homes is still pretty impressive and makes a big dent in California’s energy needs.  Think again.  There are 12.5 million households in California.   So it would take about 240 (≈12,500,000/54,000) Ivanpahs to power them all.  Ivanpah covers about 16 square kilometers.  So it would take about 3600 (= 16 x 240) square kilometers to power all these households.

Building 3600 square kilometers of mirror arrays is a big undertaking, but wouldn’t it be worth it to power the entire state of California?  The problem is that it wouldn’t power the entire state of California.  Residential power consumption is only about 20% (1/5th) of California’s total energy consumption.  Far more energy goes into commercial, industrial  and transportation needs.

If we assume vast efficiencies then we might say that it only takes 2.5 times (instead of 5 times) the residential energy consumption to run the entire state of California.  With these assumed efficiencies Ivanpah would provide the total (not just residential) energy needs for the occupants of only about 22000 (≈ 54000/2.5) homes. It would take nearly 600 (≈2.5 x 240) Ivanpahs, a whopping 9000 (≈ 3600 x 2.5) square kilometers of mirror arrays, and $1.3 trillion (≈ 2.5 x 240 x $2.2 billion) to provide the average energy needs of the entire state.

Why talk in terms of “homes?”

The use of “home” as a unit of power has a warm and fuzzy feeling to it.  I guess good and caring people are concerned about “homes,” while cold and uncaring people talk about “kilowatt-hours.”  Using “homes” as a unit of power gives the impression (intentionally?) that all the energy needs of the people living in those homes are met.  It is much more impressive to say an energy project will “power 140,000 homes” than to say it will compensate for the total energy needs for the people living in 22,000 homes.

I believe this loose use of the English language and lazy, imprecise use of physical values  is used precisely because it yields more impressive numbers.


Not much of Chinese energy is from wind or solar.

December 2, 2013

A few days ago I wrote about the pollyannish belief that “China is slowing its carbon emissions.”  An essential element of this ridiculous meme is that the Chinese are producing significant portions of their energy via wind and solar. Not true.

Consider just electricity.   Here is a breakdown of China’s installed electricity capacity by fuel type in 2011 and their electricity generation by fuel type for 2000 to 2010 from the The United Stages’ Energy Information Administration’s evaluation of China’s energy consumption (2012)…

"China's installed electricity capacity by fuel, 2011," from the US Energy Information Administration's evaluation of China's energy consumption

“China’s installed electricity capacity by fuel, 2011,” from the US Energy Information Administration’s evaluation of China’s energy consumption

"China's electricity generation by fuel type, 2000-2010" from the US's Energy Information Administration

“China’s electricity generation by fuel type, 2000-2010” from the US’s Energy Information Administration

What do these charts tell you?

These two charts are drawn from the same data set and appear next to each other in the same document.

As you can see from the top chart, 6.2% of China’s installed electricity capacity is in wind or solar.  That is over 60 gigawatts installed.  Compare that the the US’s 60 gigawatts of installed wind and 10 gigawatts of installed solar.

Alas, the top chart shows installed capacity, not actual production.  There is a little thing called the “capacity factor.”  The capacity factor is the fraction of the time that particular power source can actually produce power at its rated capacity.  For example, a one gigawatt capacity nuclear power plant will have a capacity factor of about 90%, meaning it can produce one gigawatt 90% of the time.  Wind and solar capacity factors tend to be much lower, simply because sometimes the wind doesn’t blow and the sun doesn’t shine.  The capacity factor for wind in China is 22%

The second chart shows the amount of electrical energy actually produced using the various “fuel types”.  Do you see that very, very thin yellow band along the top of the second chart?  That represents the Chinese electricity generation due to that 6.2% of installed wind and solar.  Can’t see the yellow line?  Let me blow up the last year of the chart for you…

Chinas electricity generation by fuel type blown up 3

That 6.2% of installed capacity in the form of wind and solar yields less than 1.5% of the actual energy.

China’s energy future

The Energy Information Administration document tells us…

China is the world’s second largest power generator behind the US, and net power generation was 3,965 Terawatt-hours (TWh) in 2010, up 15 percent from 2009. Nearly 80 percent of generation is from fossil fuel-fired sources, primarily coal. Both electricity generation and consumption have increased by over 50 percent since 2005, and EIA predicts total net generation will increase to 9,583 TWh by 2035, over 3 times the amount in 2010.

Wow!  three times as much as 2010, a mere 21 years from now!  Where will all this energy come from?

Again, the Energy Information Administration…

Total fossil fuels, primarily coal, currently make up nearly 79 percent of power generation and 71 percent of installed capacity. Coal and natural gas are expected to remain the dominant fuel in the power sector in the coming years. Oil-fired generation is expected to remain relatively flat in the next two decades. In 2010, China generated about 3,130 TWh from fossil fuel sources, up 11 percent annually.

Let me be clear, I am not knocking the use of wind and solar.  I have been personally working on solar energy for 17 years.  But I am knocking unrealistic expectations and quasi-religious environmentalist beliefs.  And I am not criticizing the Chinese for their increasing energy consumption.  They understand, correctly, that abundant energy is the key to prosperity.


Bjorn Lomborg on German solar subsidies

February 18, 2012

Bjorn Lomborg (The Skeptical Environmentalist and Cool It) has an interesting article (Goodnight Sunshine) describing the late realization in Germany that their massive investment in solar energy is “a massive money pit” and “a threat to the economy.”   They have subsidized 1.1 million solar photovoltaic installations to the tune of $130 billion, which provide a whopping 0.3% of the nation’s energy.

If you are worried about anthropogenic CO2 induced global warming (a.k.a. climate change, a.k.a. climate disruption), he points out

“This sizeable investment does remarkably little to counter global warming. Even with unrealistically generous assumptions, the unimpressive net effect is that solar power reduces Germany’s CO2 emissions by roughly 8 million metric tons—or about 1 percent – for the next 20 years. To put it another way: By the end of the century, Germany’s $130 billion solar panel subsidies will have postponed temperature increases by 23 hours.”

And as to “green jobs” boosting the economy of Germany…

“[E]ach job created by green-energy policies costs an average of $175,000… And many ‘green jobs’ are being exported to China, meaning that Europeans subsidize Chinese jobs, with no CO2 reductions.”

I have one large quibble with his Lomborg’s numbers.  He says

“Even with the inefficiency of current PV technology, we could meet the entire globe’s energy demand with solar panels by covering 250,000 square kilometers.”

It would really take closer to a million square kilometers.

Why a million square kilometers?

Because even good solar panels (say, rated for 150 Watts/m2) will yield much less than their name plate power for a variety of reasons.

1. A one square meter solar panel in a huge array actually requires a land area of at least 1/cos(latitude) meters.  This is the land area shaded by the one meter panel by the sun at noon.  It is larger at other times of day.

2. Extra area is required for infrastructure, such as service roads

3. The capacity factor for solar energy is about 20%.  This means that if a solar panel’s name plate says that it is 150 watts,  then on the average it will yield about 30 Watts because sometimes it is night, sometimes it is cloudy, and even during a sunny day the incident angle is optimal for only several hours before and after noon.

In Berlin, for example, where the latitude is 52 degrees, a one square meter panel in a huge array will require closer to 1.7 square meters of land.  A typical good quality one square meter panel with a name plate wattage of  150 Watts at peak averages out to about 30 watts over the course of a year because of the 20% capacity factor.  This translates to an average of about 18 Watts per square meter (30 Watts divided by 1.7 meters).

World energy consumption is about  5×1017  BTUs per year (about 1.5×1017 watt-hours/year).  That is an average power consumption of  about 1.7×1013 watts (1.5×1017 watt-hours/year divided by 8760 hours/year)

Therefore, it would take about a million square kilometers of solar arrays (1.7×1013 watts divided by 18 Watts per square meter).

This all assumes that the energy impoverished of the world are happy to stay that way.  Oh well, we all have our crosses to bear.

You can see Lomborg’s complete article here.


Comparing the Interstate Highway System to Scientific American’s “A Path to Sustainable Energy by 2030”

November 14, 2009

091111 November 09 SA cover 2In the November, 2009 issue of Scientific American, Mark Z. Jacobson and Mark A. Delucchi propose a plan to supply the world’s energy needs entirely by solar, wind and water sources by 2030. They conclude that the cost would be $100 trillion. My calculations show the cost to be more like $200 trillion.

This post dissects their comparison between the construction of the Interstate Highway System and their Energy system.


Interstate Highway System (2009 dollars):  $0.453 trillion
Jacobson’s and Delucchi’s Energy system (2009 dollars): $200 trillion

Jacobson and Delucchi say…

“Our plan calls for millions of wind turbines, water machines and solar installations. The numbers are large, but the scale is not an insurmountable hurdle; society has achieved massive transformations before… In 1956 the U.S. began building the Interstate Highway System, which after 35 years extended 47,000 miles, changing commerce and society.”

The Interstate Highway System is “largest public works program in history.” The concept was first approved by congress in 1944. But it was more than a decade until President Eisenhower signed the Federal Aid Highway Act of 1956. The plan evolved to building 42,500 miles of “super-highway” by 1975.  40,000 miles were completed by 1980.

The expected cost in 1958 was $41 billion. By 1995 the total construction cost amounted to $329 billion (in 1996 dollars). This translates into $58.5 billion 1957 dollars. That is not too far off from the original estimate.  Converting the $329 billion 1996 dollars to 2009 dollars gives $453 billion.

So if Jacobson’s and Delucchi’s estimate for the cost of their energy system is correct, then their energy plan would cost over 200 times as much ($100 trillion / $453 billion) as the Interstate Highway System to which they like to compare it.

If my calculations for the cost of their energy system are correct, then it would cost more than 400 times as much ($200 trillion / $453 billion) as the Interstate Highway System! And since they propose building their system in just 20 years, then it would be like building 20 interstate highway systems (which took about 30 years to build) every single year for twenty years.

Required surface area

Interstate Highway System – paved area: 3,500 km2
Jacobson’s and Delucchi’s Energy system (solar portion only): 500,000 km2

Composite interstate highway imageAnother interesting comparison is the amount of land required. The image at the left (click to enlarge) shows a spot check of interstate highway widths using Google Earth.  A liberal estimate of the average paved width of the Interstate Highway System is about 150 feet (about 45 meters, or 0.045 kilometers).  So, roughly speaking, the 47,000 mile (76,000 kilometer) Interstate Highway System paved over about 3,500 square kilometers ( 0.045 kilometers X 76,000 kilometers).

The area covered by solar panels in the Scientific American plan would be on the order of 500,000 square kilometers, or 150 times larger than the Interstate Highway System. (See calculated land required for Concentrated Solar, PV power plants, and rooftop solar, here)

Let’s rip up the Interstate Highway System and build a new one.

Jacobson and Delucchi claim that the expense of their energy system “is not money handed out by governments or consumers. It is an investment that is paid back through the sale of electricity and energy.” This is a soothing argument that overlooks an obvious fact: We already have a power energy system that pays for itself “through the sale of electricity and energy.”   

This is like pointing out that an Interstate Highway System would have great benefits for us, and then suggesting that we could reap those benefits by tearing down the system we have now and then rebuilding it.

It’s almost like swallowing poison so you can reap the benefits of good health after you recover.


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