August 15, 2017
END NOTES
END NOTES
CHAPTER ONE
1. http://web.stanford.edu/group/efmh/jacobson/Articles/I/USStatesWWS.pdf
“Roadmap.” Originally published in the journal Energy & Environmental Science
2. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4679003/ “Clack Evaluation.”
3. http://www.timothymaloney.net/Critique_of_100_WWS_Plan.html “Critique”
(NOTE: This paper by Tim Maloney is the basis of Roadmap to Nowhere.)
See internal footnote # 33. It refers to:
http://www.nrel.gov/docs/fy09osti/45834.pdf
Land-Use Requirements of Modern Wind Power Plants in the United States. See Page 10, Table 1, Average Area Requirements row, Total Area column: 100 hectare units (ha) = 1 km2. 34.5 ha / MWp = 0.345 km2 / MWp capacity-weighted average per NREL study in 2009.
Ibid. Chapter One End Note #1. Roadmap. See:
Table 2, row 1, column 4: 1,701,000 MWp nameplate capacity of existing plus new plants. 1,701,000 MWp X 0.345 km2 / MWp = 586,800 km2 total area for onshore wind, per NREL data (without taking into account capacity weighting of future new construction on clear flat land.)
Table 2, row 1, column 5: 3.59% existing, so 96.41% new construction. 0.9641 X 1,701,000 MW = 1,640,000 MW new construction, using 5-MW wind turbines.
When NREL made its survey in 2009, such giant 5-MW wind turbines did not exist. Using larger / taller turbines can result in an improved land density value/ This is prtt of the Roadmap’s strategy.
Note clearly that when NREL made its survey in 2009, these giant 5-MW turbines didn’t even exist. So using larger machines result in more energy production per acre, which is the Roadmap’s strategy. Hence the improved land density numbers.
Row 1, column 8: 1.5912% X 9.162e6 km2 (US total land area) = 145,800 km2 for new onshore wind construction. Anticipated new land-use density with 5-MW giant wind turbines: 145,800 km2 ÷ 1,640,000 MW = 0.089 km2 / MWp (0.0889).
So on the face of it, there is a discrepancy factor of 3.9X between NREL’s and the Roadmap’s land usage. [0.345 km2 ÷ 0.089 km2 = 3.9]
Alternatively: Total onshore wind area: 145,800 km2 ÷ 0.9641 = 151,200 km2 per Roadmap. Or 1,701,000 MW X 0.0889 km2 / MW = 151,200 km2.
NREL total wind area ÷ Roadmap total wind area: 586,800 km2 ÷ 151,200 km2 = 3.9X factor of difference.
4. Ibid. Chapter One End Note #2. Critique. See internal footnote # 22.7. Round to 160W-ac / m2 for discussion & estimation.
Also see: http://www.nrel.gov/docs/fy13osti/56290.pdf
See page 12, Sec. 4.2.1: Evaluation of PV Packing Factors. Page 13, Figure 7, Capacity-weighted average packing factor for PV projects. Fixed (mount) column: 47% packing factor (PF). 1-axis (tracking) column: 34% packing factor (PF).
Average = 40.5%, round to 40% for discussion & estimation.
Ibid. Chapter One End Note #1. Roadmap. See Table 2, row 9, column 4: 2,326,000e6 W.
2,326,000e6 W ÷ 160 W / m2 ÷ 1e6 m2 / km2 = 14.54e3 km2 total PV panel area. Land area is PV panel area ÷ PF: 14.54e3 km2 ÷ 0.40 = 36,300 km2 total land area for utility-scale PV solar, per NREL-derived data.
Table 2, row 9, column 7: 0.18973% of 9.162e6 km2 US total land area = 17,400 km2 land area for utility PV solar, per Roadmap.
NREL total PV solar area ÷ Roadmap total PV solar area: 36,300 km2 ÷ 17,400 km2 = 2.1X factor of difference.
5. www.thesolutionsproject.org/resource/50-state-visions-infographics/
Step 1. Click to download 50states_PDFs_all. In the Downloads folder, unzip and open the folder named 50states_PDFs_all.
Step 2. Double-click the Adobe Acrobat PDF icons for the 11 “great plains” states: North Dakota, South Dakota, Nebraska, Kansas, Oklahoma, Texas, Minnesota, Iowa, Missouri, Illinois, and Indiana. Upon viewing each state’s infographic, record on paper the state’s name and its percentage of Primary Energy to be provided by onshore wind. Do this on lined paper with seven drawn columns. Percent of PRI NRG from onshore wind goes in the column second from the left (column 2).
Step 3. Find each state’s Primary Energy consumption in the year 2013 by entering in your browser bar
https://knoema.com/atlas/United-States-of-America/North-Dakota/Energy-consumption
Record the large-font number at upper-left, which is North Dakota’s PRI NRG in units of billions of BTUs. In your column 3, write about 4 significant digits.
On a calculator move the decimal point 6 places to the left to express in units of quadrillion BTUs, called Quads, unit-symbol Q. Do not write it on the paper. Multiply by the conversion factor 293 TWh /Q to convert to terawatt-hour units of Primary Energy. Record in column 4.
Repeat this process 10 more times by replacing North-Dakota with the other states’ names in the browser bar. South-Dakota is hyphenated.
Step 4. Multiply each state’s 2013 Primary Energy in column 4 by the following factors to obtain its estimated Primary Energy demand in year 2050, per the Roadmap’s expectation of energy reduction. Write all 11 of the factors into column 5 before starting. These factors were obtained from the Roadmap’s Table 1, column 8, “% change in end-use power”.
ND = 0.631; SD = 0.709; NE = 0.707; KS = 0.625; OK = 0.615; TX = 0.598; MN = 0.646; IA = 0.717; MO = 0.596; IL = 0.619; IN = 0.628.
Record the multiplication results in column 6.
Step 5. Multiply each state’s estimated PRI NRG in column 6 by its onshore wind percentage from column 2. Record the result in column 7. That gives each state’s onshore wind-supplied energy in year 2050, expressed in TWh units.
Step 6. Add all 11 states’ wind consumption to obtain 3038 TWh in year 2050.
Then divide 3038 TWh by 4309 TWh to obtain 0.705, rounded to 70%. This is the portion of the nation’s onshore wind that will be located on open flat ground.
The value 4309 TWh is obtained from the Roadmap’s Table 2, onshore wind row, 30.92% in column 3. Multiply 30.92% X 13,937 TWh to obtain 4309 TWh. The value 13,937 TWh /year is the Roadmap’s standard-demand load, namely 1591 GW, converted into annual TWh energy units by multiplying X 8760 hours /yr.
Step 7. With 70% of 2050’s onshore wind capacity located on flat land where the minimum land usage value 0.089 km2 /MW pertains, that leaves 30% in harder locations where the NREL study’s 0.345 km2 /MW pertains.
Calculate the weighted average of those two values as:
0.70 X 0.089 km² + 0.30 X 0.345 km² = 0.166 km² /MW. Round to 0.17 km² /MW as the best estimate and working figure for onshore wind discussion.
Comparison to the Roadmap’s simple optimism gives a discrepancy factor of about 2X. [0.17 km² ÷ 0.089 km² = 1.9]
6. Ibid. Chapter One End Note #1. Roadmap. See frame 8 of the pdf, journal page 2098. This is the Roadmap’s Table 2, row 9, which covers Solar PV utility plants:
2,326,000e6 Wp-ac ÷ 160 Wp-ac / m2 [power rating of SunPower series E panel] = 14.5 billion m2 for utility PV panels.
Table 2, row 7 covers residential roof PV: 379,500e6 Wp-dc ÷ 186 Wp-dc / m2 [SunPower series E panel] = 2.0 billion m2 for residential PV panels.
Table 2, row 8 covers commercial roof PV: 276,500e6 Wp-dc ÷ 186 Wp-dc / m2 = 1.5 billion m2 for commercial PV panels.
All three PV solar sytems: 14.5 + 2.0 + 1.5 = 18 billion m2 of panel area.
7. Ibid. Refer to 14.5e9 m2 of utility PV panels. Rooftop PV solar: 379,500e6 W-dc (residential) + 276,500e6 W-dc (commercial) = 656,000e6 W-dc combined.
Sunpower dc power density: 158 Wp-ac / m2 ÷ 85% conversion eff. = 186 Wp-dc / m2.
Rooftop panel area: 656,000e6 Wdc ÷ 186 W / m2 = 3.5e9 m2 of rooftop PV panels. Combined utility & rooftop: 14.5e9 m2 + 3.5e9 m2 = 18.0e9 m2 total panel area.
Replaced over 40-year lifetime: 18.0e9 m2 ÷ 40 yr ÷ 365 days = 1.23e6 m2 per day.
8. Ibid. Chapter One End Note # 2. Critique. To determine the cost of the Roadmap, search in Critique for:
“Total W&S build-out cost”
“Money cost Utility PV Solar”
“Money cost Residential PV Solar”
“Money cost Commercial PV Solar”
“All three PV solar categories combined”
“Money cost Onshore Wind”
“Money cost Offshore Wind”
“Money cost CSP Solar”
9. http://www.youtube.com/watch?v=yEHf5K9AQjY at 44:55
10. Ibid. Chapter One End Note #1. Roadmap. See frame 7, journal p. 2097, bottom row, column 3:
a) 1591 GWavg total end-use power in 2050
b) 1591 GWavg X 4 hours = 6.36e12 W-hr of energy storage
Unit cost = $0.20 / W-hr
See also:
http://reneweconomy.com.au/pumped-hydro-the-forgotten-storage-solution-47248/
See the 7th paragraph:
6.36e12 W-hr X $0.20 / W h-r = $1.27 trillion construction cost for PHES
$15.2 T (from End Note #3) + $1.27 T = $16.5 Trillion
$16.5 Trillion ÷ 35 years = $471 Billion / year
11. Ibid. Chapter One End Note # 2. Critique. See internal footnote No. 65.5:
a) Average cost of KEPCO-UAE project is $22.7 billion
b) $22.7 B ÷ 5600 MWp = $4.05 / Wp
c) $4.05 ÷ 92% CF = $4.41 / Wavg for KEPCO Gen 3+ APWRs
Cost of 1,515 GWavg APWR nuclear fleet: 1,515 GW X $4.41 / W = $6.7 trillion
12.
http://innovationreform.org/wp-content/uploads/2017/07/Advanced-Nuclear-Reactors-Cost-Study.pdf
Page 10, Figure 4. Capital Cost Results. Project the rightmost two bars (MSRs) to vertical axis, at about $2000 /kW = $2 /W.
13.
14. https://www.eia.gov/analysis/studies/powerplants/capitalcost/ See:
Table 1: Supercritical coal (no Carbon Capture & Storage)
Table 2: Advanced Pulverized coal (no CCS)
15. Ibid. Chapter One End Note #1. Roadmap. Supplemental Information (SI) section
begins at Frame 28. See Table S14 on pages 66 & 67 of SI (Frames 93 & 94).
16. Ibid. Chapter One End Note #2. Critique. See internal footnote No. 66:
a) Near-term and future cost estimate of US Gen 3+ APWR = $5.53 / Wp
b) $5.53 ÷ 92% CF = $6.01 / Wavg
1,515 GWavg required APWR fleet X $6.01 / W = $9.1 trillion
17. https://www.vox.com/2016/2/29/11132930/nuclear-power-costs-us-france-korea
18. Ibid. See Figure 10 in the section “South Korea Actually Lowered Costs.” Notice that overnight construction costs have declined since 1980.
19.
20. Ibid. Chapter One End Note #12. Compare the bar heights to $4 / MWp (shown as $4,000 / kW), the approximate KEPCO price for the U.A.E. project.
21. Ibid. End Note #12, Chapter One. See Page 10, Figure 4: “Capital Cost Results.” Project the top of the two rightmost vertical bars (one of them is the ThorCon MSR) to the vertical axis. They’re both at about $2000 / kW = $2 / Watt.
Of that $2000 / kW capital cost for complete installation, about $1000 to $1200 /kW ($1.00 to $1.20 /W) is for direct construction /manufacturing cost. That cost is shown by the red portion of the vertical bars.
https://aris.iaea.org/PDF/ARISThorCon9.pdf See page 21, sub-section “Low Costs.”
See section titled “An Inconvenient Truth 2.0”
CHAPTER TWO
1. https://www.eia.gov/totalenergy/data/monthly/pdf/flow/css_2016_energy.pdf
2. https://www.eia.gov/totalenergy/data/monthly/pdf/flow /electricity.pdf
Energy consumed to generate electricity = 38.52 Quads
Gross generation of electricity = 14.69 Q
Generation efficiency = 14.69 Q ÷ 38.52 Q = 0.38
0.38 X [39% of PRI NRG] = 15% of PRI NRG
3.
http://www.goodreads.com/quotes/32944-there-are-no-passengers-on-spaceship-earth-we-are-all
4. http://www.timothymaloney.net/Critique_of_100_WWS_Plan.html Critique.
5. https://www.nrc.gov/docs/ML1034/ML103490041.pdf
Generic Aging Lessons Learned (GALL) report, Nuclear Regulatory Commission, frame 602, page X E1-2
6. https://www.wecc.biz/Reliability/2014_TEPPC_Transmission_CapCost_Report_B+V.pdf page 2-3, Table 2-1
7. http://www.pnas.org/content/114/26/6722.full
8. http://thorconpower.com/docs/domsr.pdf See page 17, 4th paragraph:
“A big shipyard . . . could easily manufacture 100 one-GW-e ThorCons per year.”
So two big shipyards = 200 GWavg annually. Therefore 1,517 GW ÷ 200 GW / year = 7.6 years.
9.
10.
http://www.americanscientist.org/issues/feature/2010/4/liquid-fluoride-thorium-reactors See p. 307, Figure 3
11. http://thorconpower.com/docs/domsr.pdf
12. https://en.wikipedia.org/wiki/Liquid_fluoride_thorium_reactor
13. http://boingboing.net/2017/07/31/nuclear-energy-is-the-safest-m.html
https://www.nextbigfuture.com/2011/03/deaths-per-twh-by-energy-source.html
See also Ibid. Chapter Two End Note #4. Critique. Search for “Land Use Utility PV Solar”, then see 15th paragraph. Also see internal FN 22.5.
16. From Chapter One End Note #5, we take the value of 1,591 GWs, minus the following:
a) Existing wind production of 21.8 GWavg in 2015 (Critique FN 67.3), with
b) Existing solar production of 4.4 GWavg in 2015 (Critique FN 67.7), with
c) Expected hydro production of 47.9 GWavg in 2050 (Roadmap, Table 2,
row 5, 3.01%).
Therefore 1,591 GWs – [21.8 GW + 4.4 GW + 47.9 GW] = 1,517 GWs
17.
https://web.stanford.edu/group/efmh/jacobson/Articles/I/USStatesWWS.pdf
Roadmap. From Table 2, row 9:
2,326,000 MWp-ac ÷ 160 Wp-ac / m2 = 14.5e9 m2 of solar panels.
To calculate PV land area divide by packing factor PF = 0.40 (40%). Obtain 36.3e9 m2 = 36,300 km2 land area; or 14,000 sq mi for utility PV solar farms.
Use the Roadmap’s assumed wind farm density of 0.089 km2 /MWp-ac:
Table 2, row 1; Wind capacity 1,701,000 MWp X 0.089 km2 / MW = 151,400 km2 land area; or 58,500 sq mi.
Combined PV & onshore wind = 14,000 + 58,500 = 72,500 sq mi for wind & PV solar.
Use the CSP land density of 0.039 km2 / MWp that describes the Andasol CSP farm in Spain (see footnote No. 86). In the Roadmap’s Table 2, rows 10 and 11, CSP capacity = 227,300 + 136,400 = 363,700 MWp. Multiply by 0.039 km2 / MW to obtain 14,200 km2, or 5,500 sq mi for utility CSP farms.
Total onshore wind and solar: 72,500 + 5,500 = 78,000 sq mi.
18. Ibid. Chapter Two End Note #11. See page 15 ff
CHAPTER THREE
1. https://www.livescience.com/15084-radioactive-decay-increases-earths-heat.html
2. http://energystoragesense.com/pumped-hydroelectric-storage-phs/
3. https://www.eia.gov/tools/faqs/faq.php?id=87&t=1
4. https://www.eia.gov/tools/faqs/faq.php?id=427&t=3
See 5th line in list: “hydro power”
CHAPTER FOUR
1. http://www.ecomodernism.org/
Download the Manifesto pdf.
2. http://www.timothymaloney.net/Critique_of_100_WWS_Plan.html Critique.
See internal footnote No. 12
4.
5. https://en.wikipedia.org/wiki/Global_warming_potential
6.
7. https://www.kcet.org/redefine/socalgas-aliso-canyon-leak-a-disaster-for-climate
37,000 tonnes methane leaked is equivalent to annual emissions of 195,000 passenger cars.
Total amount of methane leaked from Porter Ranch was 94,000 tonnes, according to CARB. By proportion, 94,000 tonnes / 37,000 t = 2.54. Multiply 195,000 cars X 2.54 = 495,000 cars. Assume 12,000 miles / yr @ 20 miles / gal; 495,000 cars X 12,000 mi / yr ÷ 20 mi / gallon = 297 million gallons of gasoline.
8. Carbon-free electric generation avoids about 405 kg CO2 emission per megawatt-hour of production, assuming that it replaces natural gas-fueled Combined Cycle Gas Turbine (CCGT) electric plants.
California wind and solar produced 20 million MW-hrs in 2013 (stated as 20 billion kW-hrs in the fourth paragraph).
https://www.forbes.com/sites/jamesconca/2014/10/02/are-california-carbon-goals-kaput/
Therefore California’s wind and solar avoided 405 kg CO2 / MW-hr X 20 million MW-hr = 8.1e9 kg CO2 = 8.1 million tonnes CO2 avoided in 2013.
Per California Air Resources Board (CARB) the Porter Ranch total emission was 94,000 tonnes of methane. At a GWP of 84X, that’s 7.9 million tonnes of CO2 equivalent (CO2-e).
7.9 million tonnes ÷ 8.1 million tonnes avoided = 98%. Therefore nearly one year’s worth of emissions benefit was wasted by Porter Ranch.
9.
See: “From Sea to Shining Sea.”
10. http://blogs.edf.org/energyexchange/2013/01/04/measuring-fugitive-methane-emissions/
See 4th paragraph.
11. Ibid. See 1st paragraph.
12. http://www.sandiegouniontribune.com/sdut-diablocanyon-naturalgas-2016jul03-story.html
13.
http://norewardisworththis.tumblr.com/post/64845798933/snl-quien-es-mas-macho-sketch-from-21719
14. http://windpower.sandia.gov/other/080983.pdf
See Page 16.
See Page 2, note 4.
15. http://onlinelibrary.wiley.com/doi/10.1029/2012GL051106/abstract
16. https://www.youtube.com/watch?v=xuttOKcTPQs
17. http://news.nationalgeographic.com/news/2004/06/0607_040607_phytoplankton.html
CHAPTER FIVE
1. http://www.timothymaloney.net/Critique_of_100_WWS_Plan.html Critique.
2. https://web.stanford.edu/group/efmh/jacobson/Articles/I/USStatesWWS.pdf
Roadmap.
See table 2, row 9: 2,326,000 MWp-ac ÷ 160 Wp-ac / m2 = 14.5e9 m2 of solar panels. To calculate PV land area, divide by packing factor PF = 0.40 (40%). Obtain 36.3e9 m2 = 36,300 km2 land area, or 14,000 sq mi for utility PV solar farms.
Use the Roadmap’s assumed wind farm density of 0.089 km2 / MWp-ac. Table 2, row 1: Wind capacity 1,701,000 MWp X 0.089 km2 / MW = 151,400 km2 land area, or 58,500 sq mi.
Combined PV & onshore wind = 14,000 + 58,500 = 72,500 sq mi for wind & PV solar.
Using the CSP land density of 0.039 km2 / MWp that describes the Andasol CSP farm in Spain:
https://en.wikipedia.org/wiki/Andasol_Solar_Power_Station
Andasol’s land area is 5.85 km2. Its nominal power rating is 150 MWp. 5.85 ÷ 150 = 0.039 km2 / MWp.)
In the Roadmap’s Table 2, rows 10 and 11, CSP capacity: 227,300 + 136,400 = 363,700 MWp. Multiply by 0.039 km2 / MW to obtain 14,200 km2; or 5,500 sq mi for utility CSP farms.
Total onshore wind and solar: 72,500 + 5,500 = 78,000 sq mi.
3. 18 billion m2 of panels ÷ 14,600 days in 40 years = 1.23 million m2 / day
4.
http://www.environmentalprogress.org/big-news/2017/6/21/are-we-headed-for-a-solar-waste-crisis
5. Ibid. Chapter 5 End Note #1 Critique. Search for “intends to ramp up our solar”.
6. Ibid. Chapter Five End Note #2. Roadmap. See the Abstract.
7. Ibid. Chapter 5 End Note #1 Critique. See internal footnotes 9 and 11.
8. Ibid. Critique. See internal footnotes 9 and 10.
9.
CHAPTER SIX
1.
https://www.gizmodo.com.au/2017/07/all-the-details-on-teslas-giant-australian- batteryt/
2. Our estimate of 77 grams of Li per kW-hr of battery storage is averaged from two sources:
A 10.8 volt (V), 8.8 amp-hour (Ah) Li-ion battery contains 7.9 grams (g) lithium.10.8 V X 8.8 coulombs / sec X 3,600 sec / h = 342e3 joules (J) energy content of battery. Conversion factor: 1 kWh = 3.6e6 J. 342e3 J X 1 kWh / 3.6e6 J = 0.095 kWh energy content of the battery. Therefore: 7.9 g Li / 0.095 kWh = 83 g lithium / kWh.
Now click on:
Refer to derivation by Saeed Kazemiabnavi: lithium content = 0.0714 kg /kWh or 71 g lithium /kWh.
Average the values 71 g and 83 g to obtain 77 g Li /kWh.
3. Ibid. Footnote #1. See 2nd paragraph:
100 MW / 129 MW-hrs refers to 129 megawatt-hours of energy storage (energy content, or energy “capacity”), with a maximum power output (discharge rate) of 100 megawatts. As usual, the word “capacity” is misused here to refer to peak power output.
129e6 W-hrs energy content X 77 g Li /1e3 W-hrs = 9.9e6 g Li, or 9.9 tonnes lithium.
4. https://en.wikipedia.org/wiki/List_of_countries_by_lithium_production
5. https://www.eia.gov/totalenergy/data/monthly/pdf/flow/css_2016_energy.pdf
6.
http://energystorage.org/energy-storage/technologies/pumped-hydroelectric-storage
7. http://thorconpower.com/costing
http://thorconpower.com/costing/bottom-line
http://thorconpower.com/docs/exec_summary.pdf
See: Frame 62, page 61.
8. http://thorconpower.com/docs/domsr.pdf
See: page 6ff
9. One cubic meter of water has mass (m) = 1000 kilograms (kg). Acceleration due to earth’s gravity (g) = 9.81 meters / second per second (9.81 m / s2). Force (F) [also called weight] = mass X acceleration = m X g. F = 1000 kg X 9.81 m / s2 = 9.81e3 newtons (N). Kinetic energy (NRG) from falling 100 meters onto hydroturbine = F X distance = 9.81e3 N X 100 m = 981e3 joules (J) per cubic meter. Conversion factor: 1 watt-hour (Wh) = 3.6e3 J.
Therefore:
981e3 J per m3 of water / 3.6e3 J /Wh = 273 Wh of kinetic NRG per m3 of water.
Ideally, 1 ESB = 917,400 m3 (with 100% efficient machinery).
273 W-hrs / m3 X 917,400 m3 = 250e6 W-hrs. Or 250 megawatt-hours per 1 ESB.
10. https://www.youtube.com/watch?v=0MJkAoA1Nek
11. The metric system is an amazing, ingenious, brilliant, and stupid-simple method of measurement based on two everyday properties of a common substance that are exactly the same all over the world: the weight and volume of water.
One cubic meter (m3) of pure H2O = one metric ton (~ 2,200 lbs) = 1,000 kilograms = 1,000 liters. And one liter = 1 kilogram (~ 2.2 lbs) = 1,000 grams = 1,000 cm3 (cubic centimeters.) And one cm3 of water = one gram, hence the word “kilogram,” which means 1,000 grams. And a tonne is a million grams.
You may have already deduced that metric linear measurements are related to the same volume of water: A meter is the length of one side of a one-tonne cube of water, and a centimeter is the length of one side of a one-gram cube of water.
Metric energy measurements are based on another thing that’s exactly the same all over the world: the force of falling water. One cubic centimeter (one gram) of water, falling for a distance of 100 meters (about 378 feet) has the energy equivalent of right around one “joule” (James Prescott Joule was a British physicist and brewer in the 1800s who figured a lot of this stuff out.)
One joule per second = one watt. (Energy used or stored over time = power. A joule is energy, a watt is power.) A million grams (one tonne) falling 100 meters per second = a million joules per second = a million watts, or one megawatt (MW). One MW for 3,600 seconds (one hour) = one MWh (megawatt-hour.)
They don’t call this a water planet for nothing.
12. https://dothemath.ucsd.edu/2011/11/pump-up-the-storage/
13. To calculate the water needed for one “grid-day” of energy: 1,591e9 W X 24 hr = 38.2e12 W-hrs. 38.2e12 W-hrs per grid-day X 1,020,000 m3 / 250e6 Wh = 156e9 m3 of fresh water = one grid-day.
https://water.usgs.gov/watuse/wuto.html
U.S. annual water use = 397 million acre-feet per year, of which 86% was fresh water, so 341 million acre-feet. Multiply by conversion factor 1.233e-6 km3 / acre-foot. Obtain 421 km3 / year, or 421e9 m3 / year
1,56e9 m3 per grid-day / 421e9 m3 water usage / year X 365 days per year = 135 days of fresh water usage for one grid-day.
14. 1,591 GWs X 24 hrs = 38.2 Terawatt hrs (trillion watt-hrs.) 38.2 trillion watts X $0.20 per W-hr = $7.64 Trillion.
CHAPTER SEVEN
1.
http://www.climatecentral.org/blogs/closer-look-at-arctic-sea-ice-melt-and-extreme-weather-15013
2. http://www.timothymaloney.net/Critique_of_100_WWS_Plan.html Critique.
Search for “115.6 tonnes / MWp”, or just “115.6”. Then search for “15,500 tonnes”. Divide that by 1,040 MW per reactor: 15,500 ÷ 1,040 = 14.9 tonnes / MWp. Then divide 115.6 t ÷ 14.9 t = 7.8.
3. Ibid. Critique. Search for “Factor of difference”.
4.
5. https://en.wikipedia.org/wiki/Cape_Wind – Controversy
6. https://web.stanford.edu/group/efmh/jacobson/Articles/I/USStatesWWS.pdf
Roadmap.
See Frame 8, journal page 2098, Table 2. Use columns 2 and 6, for rows 1, 2, 9 and 10.
7. http://www.meteo.mcgill.ca/~huardda/articles/greene10.pdf
See pages 1594, 1595, 1599.
8. https://www.nasa.gov/topics/earth/features/warmingpoles.html
CHAPTER EIGHT
1.
https://web.stanford.edu/group/efmh/jacobson/Articles/I/USStatesWWS.pdf
Roadmap.
See Frame 5, journal page 2095
2. http://www.timothymaloney.net/Critique_of_100_WWS_Plan.html
Critique.
Search for “Figure C”; caption.
3. Ibid. Critique.
Search for Figure C. Observe Figure C call-out of 30,000 m3 water / hour. That means 24 hours / day, forever. 30,000 m3 / hr X 8760 hr / year = 263e6 m3 /yr.
Divide by U.S. annual fresh water usage of 421e9 m3 / year: 263e6 ÷ 421e9 = 0.000 62. Multiply 0.000 62 X 8760 hours /yr = 5.4 hours.
4. https://www.youtube.com/watch?v=VgQQOZkdxag
5.
CHAPTER NINE
1.
https://web.stanford.edu/group/efmh/jacobson/Articles/I/USStatesWWS.pdf
Roadmap.
Table 2, row 12, column 3
2. Ibid. Table 2, row 11, column 3
3. http://www.timothymaloney.net/Critique_of_100_WWS_Plan.html
Critique.
See internal footnote 67.5
4. Ibid. Chapter Nine End Note #1 Roadmap. Table 2, row 4, columns 3 and 5 for already installed geothermal.
5.
Germany’s installed (peak) wind capacity, versus actual (average) production:
https://wryheat.wordpress.com/2015/02/12/german-wind-power-fails-a-cautionary-tale/
6. http://www.whoi.edu/page.do?pid=83397&tid=3622&cid=94989
http://www.timothymaloney.net/Pacific_Ocean_damaged_by_Fukushima.html
Search for “It’s all the same”
https://www.forbes.com/sites/jamesconca/2014/09/05/germans-boared-with-chernobyl-radiation/
CHAPTER TEN
1.
https://carboncounter.wordpress.com/2015/08/11/germany-will-never-run-on-solar-power-here-is-why/
2. http://www.pnas.org/content/114/26/6722.full
3. http://www.pnas.org/content/112/49/15060
Frame 5, page 15064, Figure 4B.
4. http://search.usa.gov/search?utf8=%E2%9C%93&affiliate=eia.doe.gov&query=existcapacity_annual.xls
Then click on www.eia.gov . Open or Save the offered file. A spreadsheet doc will come up. Scroll to line 38,854 for the year 2015: “Hydroelectric.” Go to column “Nameplate Capacity”: 78,957 MW (79.0 GW).
5.
See paragraph 26 (but the entire article is worth reading, too.)
6. http://www.postcarbon.org/controversy-explodes-over-renewable-energy/
7. In 2015 there were 8,002 dedicated electricity-producing facilities in the U.S.
http://www.eia.gov/electricity/annual/html/epa_04_01.html
See: “Total Sectors” section, row 2015.
8.
https://www.hbr.org/2017/04/the-3-stages-of-a-country-embracing-renewable-energy
10th paragraph: “… , grid operators frequently have to intervene to keep the electricity grid in balance. For example, interventions in Germany’s largest transmission grid operated by private company TenneT increased from fewer than 10 interventions per year in 2003 to almost 1,400 interventions in 2015.”
13th paragraph: “. . . demand-response . . . temporarily switch off part of their electricity consumption—increasing the elasticity of demand to keep the grid balanced.”
“Germany’s push toward renewable energy is causing so many drops and surges from wind and solar power that more utilities than ever are receiving money from the grids to help stabilize the country’s electricity network.
“Twenty power companies . . . add or cut electricity within seconds to keep the power system stable, double the number in September, according to data from the nation’s four grid operators.
“Germany’s drive to almost double power output from renewables by 2035 has seen one operator reporting five times as many potential disruptions . . . “
CHAPTER ELEVEN
1.
See page 2.
2. http://www.timothymaloney.net/Critique_of_100_WWS_Plan.html Critique.
See internal footnotes 12 and 27
3. ttps://web.stanford.edu/group/efmh/jacobson/Articles/I/USStatesWWS.pdf
Roadmap.
See Table 2, footnote d.
www.pnas.org/content/112/49/15060
See Table 2, footnote c.
4. Ibid. Chapter Eleven End Note #2. Critique.
Search for “About 160 W”. See internal footnote 22.3.
5. Ibid. Chapter Eleven End Note #2. Critique.
Search for “US Solar PV System Cost Benchmarks”; then search for “28%”, then “23%”
6. https://en.wikipedia.org/wiki/Moore%27s_law
https://www.greentechmedia.com/articles/read/why-moores-law-doesnt-apply-to-clean-technologies
7. http://www.nrel.gov/docs/fy16osti/67142.pdf
See graph on page 8.
8. Ibid.
See graph on page 42: “Modeled Impacts of Module Efficiency on Total System Costs, 2016.”
9.
https://web.stanford.edu/group/efmh/jacobson/Articles/I/USStatesWWS.pdf
Roadmap
Table 2, row 9, column 7 states 0.18973% as the portion of US land required for utility solar PV farms.
0.18973% X 9.162e6 km2 total US area = 17,380 km2 land area required for utility PV, asserted by the Roadmap.
Referring to Table 2’s column 4, using 160 W / m2 SunPower PV panels specified by the Roadmap, the total panel area (not land area), is 2,326,000e6 W ÷ 160 W /square meter = 14.54e9 square meters of total panel area.
At U.S. average packing factor of 40%, the total land area required = total panel area ÷ 0.40: 14.54e9 m2 ÷ 0.40 = 36.35e9 m2 = 36,350 km2. This is 0.397% of total U.S. land.
Therefore the NREL / SunPower-derived land requirement for utility-scale PV solar is 2.1X greater than the Roadmap’s assertion. [36,350 km2 ÷ 17,380 km2 = 2.1. Also 0.397% ÷ 0.18973% = 2.1.]
The Critique’s treatment of this issue can be found by searching for “would occupy only 37,100”. The discrepancy between the Critique’s 37,100 km2 of land and 36,350 km2 calculated here is due to the Critique’s rounding of land density values to just 2 significant figures, namely 0.029 and 0.016 km2 /MW.
10. Ibid. Chapter Eleven End Notes #7
See Page 8, “Overall Model Results.” Utility Scale PV cost values are at the right.
2016 fixed-tilt cost = $1.42 per dc watt. For ac divide by dc-to-ac conversion factor 0.83, the value assumed by NREL for ground-mounted PV facilities.
$1.42 /Wdc ÷ 0.83 = $1.71 / W-ac (for ground-mounted fixed-tilt in 2016)
Page 45, Conclusions (1), for single-axis tracking mount;
$1.49 / W-dc ÷ 0.83 = $1.79 / W-ac (for single-axis tracking in 2016)
$1.71 for fixed-tilt and $1.79 for tracking-mount, per ac watt. Combined average $1.75 per ac watt.
11. Ibid. Chapter Eleven End Note #7.
Page 8, Overall Model Results, Utility Scale PV: all dollar values are per dc watt.
$1.42 for fixed-tilt and $1.49 for tracking-mount. Combined average $1.45 per dc watt.
See page 36, Utility Scale PV, Modeling Inputs. All dollar values are per dc watt:
Module Price:
$0.64 ÷ $1.42 = 45% for fixed mount
$0.64 ÷ $1.49 = 43% for tracking mount
44% average, for PV module cost portion.
Inverter Price:
$0.09 ÷ $1.42 = 6.3% for fixed mount
$0.10 ÷ $1.49 = 6.7% for tracking mount
6.5% average, for inverter cost portion; rounded to 7% in text.
Installation Labor:
http://www.nrel.gov/docs/fy15osti/64746.pdf
For 2015 installations, see page 29, Figure 21:
$0.16 /Wdc for fixed mount
$0.22 /Wdc for tracking mount
$0.19 /Wdc average, labor cost portion in 2015.
Labor cost declined by about one-third from 2015 to 2016. See NREL 2016 report (Ibid.), page 8, Utility Scale PV, at far right. Compare orange-color segments in the bar graphs for those two years. By comparison, estimate that $0.19 declined to about $0.13 /Wdc. $0.13 /Wdc ÷ $1.45 /Wdc = 9.0%, labor cost portion in 2016.
44% (module cost) + 6.5% (inverter cost) + 9% (labor cost) = 60% of initial cost of utility PV solar.
12. Assuming that initial labor cost of 9% divides as 5% for panels and 4% for inverters:
1 PV panel replacement = 44% module + 5% labor = 49%
3 inverter replacements = 3 X (6.5% parts + 4% labor) = 31%
Lifetime replacement cost = 49% + 31% = 80%.
Lifetime cost factor = 1.80X.
13. 2,326,000 million watts (MWs, or megawatts) X $1.75 /W = $4.1 trillion for initial installation at 2016 cost.
14. $4.1 trillion X 1.80 = $7.4 trillion lifetime cost, before NREL future discount.
16. Ibid. Chapter Eleven End Note #7
Page 8, Overall Model Results: residential cost values are at the left. All values expressed in dc watts. (vertical scale factor = $0.11 /millimeter)
$2.93 / Wdc for residential solar in 2016.
17. 379,500 million watts X $2.93 / W = $1.1 trillion for initial installation at 2016 cost.
18. Ibid. Chapter Eleven End Note #7
p. 25, Residential PV. Modeling Inputs and Assumptions:
module + string inverter = $0.64 + $0.16 = $0.80 /Wdc
module: $0.64 ÷ $2.93 = 22%
inverter: $0.16 ÷ $2.93 = 6%
p. 8, bar graph; Scale orange segment for labor: 2.7 mm. Vertical scale factor = $0.11 per mm; installation labor = 2.7 mm X $0.11 / mm = $0.30 /Wdc
Labor: $0.30 ÷ $2.93 = 10%
module + string inverter + labor = $0.80 + $0.30 = $1.10 / W-dc
$1.10 ÷ $2.93 = 38%
19. Assuming that initial labor cost of 10% divides as 5% for panels and 5% for inverters:
1 PV panel replacement = 22% module + 5% labor = 27%
4 inverter replacements = 4 X (6% parts + 5% labor) = 44%
Lifetime replacement cost = 27% + 44% = 71%.
Lifetime cost factor = 1.71X.
20. $1.11 trillion X 1.71 = $1.90 trillion lifetime cost, before NREL future discount.
21. Ibid. Chapter Eleven End Note #7. See Page 8, section on “Overall Model Results.” Commercial cost values are in the center. Values are expressed in dc watts. (Vertical scale factor = $0.11 /millimeter.) $2.13 / Wdc for commercial rooftop solar in 2016.
22. Ibid. Chapter Eleven End Note #7
See Page 31, Commercial PV “Modeling Inputs and Assumptions”:
module + inverter = $0.64 + $0.13 = $0.77 /W-dc
module: $0.64 ÷ $2.13 = 30%
inverter: $0.13 ÷ $2.13 = 6.1%
See also Page 8, bar graph. Commercial cost values are in the center. If we scale the orange segment for labor, we get 1.8 mm. At $0.11 per mm. Installation labor = 1.8 mm X $0.11 / mm = $0.20 / W-dc. Labor: $0.20 ÷ $2.13 = 9.4%
module + inverter + labor = $0.64 + $0.13 + $0.20 = $0.97 / W-dc
$0.97 ÷ $2.13 = 46%
23. Assuming the initial labor cost of 9.4% divides evenly as 4.7% for panels and 4.7% for inverters:
1 PV panel replacement = 30% module + 4.7% labor = 34.7%
4 inverter replacements = 4 X (6.1% parts + 4.7% labor) = 43.2%
Lifetime replacement cost = 34.7% + 43.2% = 78%
Cost factor = 1.78X.
24. 276,500 million watts X $2.13 / W = $590 billion for initial installation at 2016 cost.
25. $590 billion X 1.78 = $1050 billion lifetime cost before NREL future discount.
26. Utility + residential + commercial = $5.3 T + $1.5 T + $0.8 T = $7.6 T, if NREL’s future discount projection is borne out.
27. Ibid. Chapter Eleven End Note #2. Critique.
Search for “factor of 16.9”. Refer to internal footnote # 13. Then refer to “factor of 58” in internal footnote # 14.
28. See the “Material Requirements” chapter in “The Non-Solutions Project” by Mathijs Becker:
https://www.amazon.com/non-solutions-project-Mathijs-Beckers/dp/1537673807
29. http://www.bbc.com/future/story/20150402-the-worst-place-on-earth
30. A CSP farm’s capacity factor is sometimes specified in the 40% or 50% range. That’s an unrealistic range for actual solar insolation anywhere in the entire world, even in the sunniest locations. Such a spurious capacity factor is conjured by the solar industry with the following accounting gimmick:
Only a portion of a CSP farm supplies immediate electric energy to the grid. The rest of the farm’s curved mirrors put heat energy directly into a pipe of molten salt, and the hot salt is stored in insulated tanks for later use.
But instead of adding up all the energy (electric + heat) produced by the entire farm, the solar operator only counts the “immediate electric energy” portion of the farm as the total peak power rating for the entire farm.
The farm’s electric generating equipment and steam turbine are sized to handle just the amount of power produced by the immediate electric energy mirrors, and not the entire solar field, meaning the farm’s entire collection of mirrors. The industry has coined the innocuous term “solar multiple” for this accounting gimmick.
Solar multiple is the ratio of all the mirrors in the solar field to the mirorrs that are producing immediate electric energy. For example, a solar multiple of 1.5 means that in a 150-mirror CSP farm, 100 mirrors are counted and 50 mirrors aren’t. This reduces the farm’s declared peak-power rating. However, the material use and dollar cost to build the entire field relates to the entire solar field, and not just the counted mirrors.
As described, the 50 uncounted mirrors store the thermal energy that’s intended to be used after sundown. The accounting trick makes it look on paper like the additional electric output obtained from the stored (and uncounted) thermal energy seems to be coming from less infrastructure than it really is.
This enables the industry to (falsely) quote a greater capacity factor of the CSP farm, for advertising and PR purposes.
31. https://en.wikipedia.org/wiki/Andasol_Solar_Power_Station
Andasol’s land area is 5.85 km2. Its nominal power rating is 150 MWp. 5.85 ÷ 150 = 0.039 km2 /MWp.
32. Ibid. Chapter Eleven End Note #2. Critique.
Search for “$5.94”. Refer to internal footnote 46.
CHAPTER TWELVE
1. https://web.stanford.edu/group/efmh/jacobson/Articles/I/USStatesWWS.pdf Roadmap.
See table 2, row 1, columns 2, 6 and 8. U.S. land area is taken to be 9.162e6 km2.
1.5912% X 9.162e6 km2 = 145,800 km2; 5 MW X 328,000 turbines = 1.64e6 MW;
145,800 km2 ÷ 1.64e6 MW = 0.089 km2 / MWp assumed land density for wind.
2. On 10/20/16 11:32 PM, Timothy Maloney sent this message:
Dear Dr. Jacobson,
I am writing an article on renewable energy and need some clarification.
For onshore wind the 100% clean and renewable WWS all-sector energy roadmaps for the 50 united States shows 1.59% of US land area needed for spacing of new plants /devices. I take this to mean the entire area of a wind farm, what the National Renewable Energy Laboratory defines as Total Wind Plant Area in their 2009 technical report – Land-Use Requirements of Modern Wind Power Plants in the United States.
NREL defines Total Wind Plant Area as “the total area of a wind power plant consisting of the area within a perimeter surrounding all the turbines in the project”. [p.4, Sec. 2.2]
172 large wind projects were evaluated in the NREL study, obtaining a clear specification of the Total Wind Plant Area for 161 of them [p.10, Table 1]. Their combined Total Area was 8778.9 km², with combined generating capacity of 25,438 MWac, giving an Average Area Requirement of 34.5 ha /MW, or 0.0345 km² /MW, shown at lower right in that table.
The 100% WWS Roadmap, Table 2, states a target value of 1,701,000 MW, with 3.59% already built as of 2013. New buildout would therefore be 1,640,000 MW.
With NREL’s land-usage for actually existing large wind farms at 0.0345 km² /MW, the new land area required would be 565,800 km². That land area represents 6.18% of all US land, if Alaska is counted. This is about 4X greater than the 1.59% value for onshore wind in Table 2 of the Roadmap.
Perhaps the word “spacing” in Table 2 does not really refer to the Total Area occupied by large wind farms built in the US. Perhaps it refers instead to a theoretical model for flat land only, assuming a rectangular field with a turbine array spaced about 3 to 5 blade-diameters apart “sideways,” and 10 diameters apart in the direction of prevailing wind.
Under that assumption, analysis models anticipate land usage of 0.13 to 0.20 km² /MW [p.15 of the NREL report]. The center value of that predicted range, 0.165 km² /MW, would yield new land requirement of 270,600 km², or 2.95% of total US area. Even this idealization is substantially greater than the 1.59% of US land area specified in Table 2.
Could you help me reconcile these discrepancies?
Thank you in advance.
Timothy Maloney
On 10/21/16 1:17 AM, Mark Z. Jacobson replied:
Dear Timothy,
Yes, I will address this below. Also, I checked out your “Critique” of our U.S. plan, and while I am flattered you have taken such an interest, I would suggest you go into the spreadsheets more to see exactly how things are calculated. For example, you claim that the U.S. average capacity factor of wind and solar applied to our generation capacity give a slight underestimate of our annual power output but you omit the fact that we are including offshore wind in our 2050 mix (none of which existed in the U.S. at the time of the report), and CFs are higher for offshore wind than onshore, and you averaged wind and solar CFs and different types of solar CFS, then multiplied an average number by a total capacity rather than multiplying individual CFs by individual capacities and summing the results. Also, you used recent values rather than 2050 values, which we use.
With regard to wind turbine spacing areas, we use the standard metric for wind turbine area requirement Area (km2) per turbine = aD x bD where D is turbine diameter (km), and a and b are constants representing the sidestream and upstream distance between turbines in an area. For onshore turbines, we used a=4, b=7 and for offshore, a=5 and b=10. For the 5-MW, D=126 m turbine we used, these translate into 0.44 km^2/turbine (0.089 km^2/MW) and 0.79 km^2/turbine (0.159 km^2/MW), respectively.
A recent study that will be published shortly by an independent group analyzing the spacing of more than 1000 operating turbines covering 44 onshore and offshore wind farms around the world found that the mean distance between turbine towers was 4.2D, giving an approximate mean area of turbines as A = 4.2D x 4.2D = 17.6 D^2, which is much less than what we used (28 D^2 and 50 D^2).
In other words, the spacing areas we estimated are larger than spacing based on real wind farm data (thus our results are conservative), which is opposite from the conclusion you draw from the NREL report.
There are three reasons for this.
1) NREL does not provide any calculation of actual average distances between turbines towers, which is the relevant method of performing this calculation because the reason turbines are spaced is to avoid interference of the wake of one turbine with the next. It is irrelevant to know the irregular outside perimeter of a property based on project applications (which is what NREL used), particularly since the outside may be far away from the last turbine actually installed or could lie in a creek bed far away from any turbines.
2) The NREL report acknowledges on page 15 that their method of calculation “Wind Plant Area” results in overestimates and gives several examples why.
3) On page 4 of the NREL report, they further acknowledge that the Wind Plant Area is “subjective in nature” and “the total area of a wind power plant could have a number of definitions.” In their case, they define it based on project applications, which results in several of the overestimates given in (2) above.
On the other hand, the method based on data I described above relies on analyzing actual distances between turbine towers.
In sum, I believe our estimates overestimate rather than underestimate spacing area requirements based on real data.
This result is common sense as well, particularly as we go toward 1.7 million turbines in the U.S. Wind farm operators have an incentive to squeeze turbines as close together as possible to minimize transmission costs and land impacts, sacrificing some loss in capacity factor due to more interference.
Sincerely,
Mark Jacobson
3. http://www.timothymaloney.net/Critique_of_100_WWS_Plan.html Critique.
Search for “NREL’s 0.029 value becomes 0.016”.
4. Ibid. Chapter Twelve End Note #1. Roadmap. Table 2, row 1, column 3.
5. Ibid. Chapter Twelve End Note #2. Critique. See internal footnote 37.
6. Ibid. See internal footnote 41.
7. Ibid. Chapter Twelve End Note #1. Roadmap. Table 2, row 2, column 3.
8. Ibid. Chapter Twelve End Note #2. Critique.
See internal footnote No. 40. See also:
http://onlinelibrary.wiley.com/doi/10.1002/we.v20.2/issuetoc
Wind Energy Feb. 2017, Volume 20, Issue 2, pages pages 361-378.
FINAL REMARKS
1. http://www.thesciencecouncil.com/pdfs/P4TP4U.pdf
See Chapter Seven.
2.
https://www.theguardian.com/environment/2017/jul/31/paris-climate-deal-2c- warming-study
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