Which car would you rather drive?

Which of these two cars would you rather drive… 

Smart Car

Edison 2

The top car is the mis-named “Smart Car” from Daimler AG.  For somewhere between $12,000 and $20,000 you get a vehicle that carries two people gets 33 mpg city and 41 mpg highway for a combined 36 mpg.  That’s almost 85% of the fuel economy of a Honda Civic – fifteen years ago.    

The bottom picture is of the  Automotive XPrize winner,  the Edison2.  The Edison2 seats four, gets over 100 mpg, has a top speed of of 110 mph and a range of over 600 miles on a single tank of gas.  It will travel 50 mph on a mere 3.5 horsepower. and will go from 0 to 60 in less than 10 seconds.  And it would cost half as much as a Chevy Volt.   Edison2 is headed up by Oliver Kuttner, and according to consumer reports Kuttner says the Edison2

has plans for a car that is closer to being production ready, with bodywork that sounds more substantial. Should it progress to production, the car could be offered in the $20,000 range.

The call is yours, which would you rather drive? 

Scientific American’s “A Path to Sustainable Energy by 2030:” the Cost

The cover story of the November issue of Scientific American, A Path to Sustainable Energy by 2030,” by Mark Z. Jacobson and Mark A. Delucchi  promises a path to a “sustainable future” for the whole world in just 20 years. They define “sustainable” as a world where all energy sources are derived from water, wind and solar. Nuclear need not apply.

The article had a few words about the cost, but much was left out.  Jacobson and Delucchi conclude that their grand plan will cost about $100 trillion dollars.  I found this ridiculously large sum to be too low!  My rough calculations yields a cost of $200 trillion!

This post is an attempt to fill in a few blanks.

I will accept the authors’ mix of energy sources, apply some capacity factor estimates for each source, throw in an estimate of the land required for some sources, and estimate the installation cost per Watt for each source. Since all of these numbers are debatable, I provide references for most of them. But some of the numbers are simply my estimates. Also, I consider only installation costs.  I do not consider the additional costs of operation and maintainance, which may considerable.

Another point, the authors say that the US Energy Information Administration projects the world power requirement for 2030 would be 16.9 TW to accomodate population increase and rising living standards. By my reading, the Energy Information Administration’s estimate is actually 22.6 TW by 2030¹³.  Nevertheless, Jacobson and Delucchi base their plan on only 11.5 TW, with an assumption that a power system based entirely on electrification would be much more efficient.  I will go along with their estimate of 11.5 TW for the sake of argument.

Here are my numbers

(click on image to get larger view)…

 

The numbers that I have placed in the blue columns are open to debate, but I am fairly confident of the capacity factors.  The capacity factor for concentrated solar power, with energy storage, such as molten salt, can vary depending on interpretation.  If energy is drawn from storage at night, then the capacity factor could be argued to be higher.  On the other hand, it would result in greater collection area, collection equipment and expense.   Note that using my estimates for capacity factors, the “total real power” works out to 12.03 TW, close to Jacobson’s and Delucchi’s 11.5 TW.

The dollars per installed watt is where I would expect the greatest argument.  For example, Jacobson and Delucchi call for 1.7 billion 3000 watt rooftop PV systems.  That is residential size, on the order of 300 square feet.  You can find offers for residential systems at much lower rates than $8 per watt installed.  But this is because of rebates and incentives.  Rebates and incentives only work when a small fraction of the population takes advantage of them.  If every residence must install a photovoltaic system, there is no way to pass the cost on to your neighbors.  Click on the chart on the left, from Lawrence Berkeley National Laboratory: of all the states listed, only one comes in at under $8 per installed watt for systems under 10 kilowatts, and half of the remaining come in at over $9.

Wouldn’t prices fall as technology advances?  Not necessarily.  Look at the cost to install wind facilities – it has been increasing since the early 2000s. A large part of the installed price for wind is the cost of the wind turbine itself.  Click on this graph showing the price of wind turbines per kilowatt capacity.  This increasing trend will likely continue if demand is artificially pushed up by a grandiose plan to install millions more wind turbines beyond what are called for by the free-market.

Expect to see the same effect for photovoltaic prices.  While the cost of photovoltaic power has been slowly falling, the demand (as a fraction of the total energy market) has been miniscule.  Jacobson and Delucchi call for 17 TW of photovoltaic power (5 TW from rooftop PV and 12 TW from PV power plants) by 2030.  Compare that to the what is already installed in Europe, the world’s biggest marked for PV: 0.0095 TW.  Achieving Jacobson’s and Delucchi’s desired level would require an orders or magnitude demand increase.  This is likely to lead to higher prices, not lower.  For my calculations I am staying with today’s costs for photovoltaics.

Some perspective

We have started using the word “trillion” when talking about government expenditures.  Soon we may become numb to that word, as we have already become numb to “million” and “billion.”  My estimate for the cost of Jacobson’s and Delucchi’s system comes out to about $210 trillion.  So how much is $210 trillion dollars?

It is approximately 100 times the $2.157 trillion of the total United States government receipts of 2009 (see documentation from the Government Printing Office) . 

It is about 15 times the GDP of the United States.

$210 trillion dollars is about 11 times the yearly revenue of all the national government budgets in the world!  You can confirm this by adding all the entries in the revenue column in the Wikipedia “Government Budget by Country.”

What about just the United States?

Jacobson and Delucchi calculate that with their system the US energy demand with be 1.8 TW 2030.  Keep in mind that the demand today is already 2.8 TW.  If we accept their estimate of 1.8 TW, then that  is about 16% of their estimated world demand of 11.5 TW for 2030.  So roughly speaking, the US share of the cost would be 16% of $210 trillion, or about $34 trillion.  That is 16 times the total United States government receipts of 2009. 

Doesn’t seem to likely to work, does it?

I know that Jacobson and Delucchi don’t like nukes.  But the Advanced Boiling Water Reactor price of under $2 per installed watt sure sounds attractive to me now.  Just a thought.

Update 11/14/2009

Jacobson and Delucchi compared their scheme to the building of the interstate highway system.  See here for are realistic comparison.

Notes

1) Capacity factor of wind power realized values vs. estimates, Nicolas Boccard, Energy Policy 37(2009)2679–2688
2)  http://www.oceanrenewable.com/wp-content/uploads/2009/05/power-and-energy-from-the-ocean-waves-and-tides.pdf
3)  Fridleifsson,, Ingvar B.,  et. al.,  The possible role and contribution of geothermal energy to the mitigation of climate change. (get copy here)
4)  http://en.wikipedia.org/wiki/Hydroelectricity
5)  Tracking the Sun II, page 19 , Lawrence Berkeley National Laboratory, http://eetd.lbl.gov/ea/emp/reports/lbnl-2674e.pdf
6)  Projecting the Impact of State Portfolio Standards on Solar installations, California Energy commission, http://www.cleanenergystates.org/library/ca/CEC_wiser_solar_estimates_0205.pdf
7)  David MacKay – “Sustainable Energy – Without the Hot Air” http://www.withouthotair.com/download.html
8).  64MW/400acres = 40MW/km2 http://www.chiefengineer.org/content/content_display.cfm/seqnumber_content/3070.htm
9)  http://www.windustry.org/how-much-do-wind-turbines-cost
10)  I have chosen a low cost because most hydroelectric has already been developed.
11) 280 MW for $1 billion, http://www.tucsoncitizen.com/ss/related/77596
12) Based on my personal experience as a Scientist working on photovoltaics for 14 years at the National Renewable Energy Laboratory.  This number varies according to insolaton, latitude, temperature, etc.
13)  The EIA predicts a need for 678 quadrillion (6.78 x 10¹⁷) BTUs of yearly world energy use by 2030.  One BTU is the same as 2.9307 x 10^-4  kiloWatt hours.   So, (6.78 x 10¹⁷ BTU) x (2.9307 x 10^-4  kWhr / BTU) = 1.98 x 10¹⁴ kWhr.    One year is 8.76 x 10³ hours.  So the required world power would be given by:  (1.98 x 10¹⁴ kWhr) / (8.76 x 10³ hr) = 2.26 x 10¹⁰ kW = 22.6  TW.

Bad professors, BAD. The truth about “Eat the Dog”

Guest post from Cocoa the dog

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I am told humans are smart, but sometimes I wonder.   I was born back in ’02, and I have learned a trick or two in my 49 years.  But this old dog will never play the kind of trick that Brenda and Robert Vale are playing.  They are off by a factor of 20 when comparing the energy to power an SUV with the energy to power a dog.

Brenda and Robert Vale are professors at Victoria University in Wellington, New Zealand.  They are either complete mathematical boneheads, or they have simply realized that in today’s world there is no limit to the outrageous claims that they can peddle to other completely credulous humans.  They claim in their book “Time to Eat the Dog: The real guide to sustainable living” that I am an energy hdogs – worse than a gas guzzling SUV.  Here is their (il)logic, as reported in the New Zealand Dominion Post…

The couple have assessed the carbon emissions created by popular pets, taking into account the ingredients of pet food and the land needed to create them.

“A lot of people worry about having SUVs but they don’t worry about having Alsatians and what we are saying is, well, maybe you should be because the environmental impact … is comparable.”

In a study published in New Scientist, they calculated a medium dog eats 164 kilograms of meat and 95kg of cereals every year. It takes 43.3 square metres of land to produce 1kg of chicken a year. This means it takes 0.84 hectares to feed Fido.

They compared this with the footprint of a Toyota Land Cruiser, driven 10,000 kilometers a year, which uses 55.1 gigajoules (the energy used to build and fuel it). One hectare of land can produce 135 gigajoules a year, which means the vehicle’s eco-footprint is 0.41ha – less than half of the dog’s.

Let me help my two-legged friends with their calculations.

Let’s compare the amount of land needed to generate enough biofuel to drive a Toyota Land Cruiser 10,000 km, to the amount of land required to feed a dog.  Let’s compare kibbles to kibbles.  In the case of the Land Cruiser grain may be converted to ethanol to power the vehicle.  Similarly, grain can be fed to animals to yield meat, which can be fed to the dog. 

Land Cruiser

My farm animal friends tell me that corn is the best grain for making ethanol.  In the US, where they grow a lot of corn, they got 371 bushels of corn per hectare in 2007.*  Each bushel of corn gives about 2.7 gallons of ethanol according to the USDA.  So that means each hectare of corn yields about 1000 gallons of ethanol.**

The humans at Toyota say that the Land Cruiser gets 13 miles (20.8 kilometers) per gallon in the city and 18 miles (28.8 kilometers) per gallon on the highway.  But that is when it runs on gasoline.  The energy content of gasoline is 115,000 BTU/gallon.  But for ethanol it is only 75,700 BTU/gallon.  So it takes about 50% more ethanol to get the same energy.***  That is, the Land Cruiser would only get 8.6 miles (13.8 kilometers) per gallon of ethanol in the city and 11.8 miles (18.9 kilometers) per gallon of ethanol on the highway.****  Let’s average it and call it 10.2 miles (16.3 kilometers) per gallon of ethanol for the Land Cruiser.

So it takes 613 gallons of ethanol to drive the Land Cruiser 10,000 kilometers.  That translates into 0.61 hectares of corn land. *****

Feeding a dog

Remember, a hectare of corn gave 371 bushels of corn in 2007.  A bushel of corn weighs 56 pounds (25.5 kilograms).  That is 20,776 pounds (9,441 kilograms) of corn per hectare.⁺

If you want to convert that corn into chicken meat, as the professors suggest, then according to the Agricultural branch of the Australia’s Department of Primary Industries, the conversion factor is about two kilograms of chicken feed to one kilogram of chicken liveweight.   That means that a hectare of corn would give about 10,388 pounds (4,722 kilograms) of chicken liveweight.  Dogs are not as fussy as humans, but even we don’t eat the feathers. We would only eat about 2/3 of the bird liveweight.  That fetches 6925 pounds (3147 kilograms) of edible meat per hectare.⁺⁺

According to the boneheaded professors, a typical dog eats 164 kilograms of meat per year.  (I have a pretty good life – but I can tell you I don’t eat nearly that much. But I’ll play along anyway.)  That would require 0.052 hectares to produce.⁺⁺⁺  They say that we also eat another 95 kilograms of cereals each year – or another 0.01 hectares worth of corn.⁺⁺⁺⁺  That sniffs out to 0.062 hectares worth of land to feed an overfed dog.

Conclusion

0.61 hectares to feed the soulless Toyota Land Cruiser.

0.062 hectares to feed your best friend.

That’s 10 times as much for the Land Cruiser than for me.  I could have sworn the professors said the dog required twice as much land as the Land Cruiser.  They were only off by a factor of 20.

Bad professors, BAD.  Don’t make me rub your nose in it.

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* (151.1 bushels / hectare) x (2.46 acres / hectare) = 371 bushels per hectare.

** (371 bushels)  x  (2.7 gallons/bushel) = 1006 gallons

*** 115,000 BTU  /  75,700 BTU  =  1.52

**** (13 miles / gallon) / 1.52  =  8.6 miles / gallon = 13.8 kilometers / gallon
**** (18 miles / gallon) / 1.52  =  11.8 miles / gallon = 18.9 kilometers / gallon

***** 10,000 kilometers / (16.3 kilometers / gallon) / (1002 gallons/ hectare) = 0.61 hectares

⁺ (371 bushels/hectare) x (56 pounds/bushel) = (20,776 pounds/hectare) = (9443 kilograms/hectare)

⁺⁺ (20,776 pounds/hectare) x (1/2)  x  (2/3) = (6925 pounds/hectare) = (3147 kilograms/hectare)

⁺⁺⁺ (164 kg of meat/dog) / (3147 kg of meat/hectare) = (0.052 hectares/dog)

⁺⁺⁺⁺ (95 kg of corn/dog) / (9,441 kg of corn/hectare) = (0.01 hectares/dog)