Energy consumption: fossil fuels and the growth issue 1
Daily energy consumption per capita in 2005 WORLD United Arab Emirates Kuwait Canada USA Australia Netherlands Sweden Saudi Arabia Russian Federation France Japan Germany Switzerland United Kingdom Spain Denmark Venezuela Poland Mexico China Brazil India Philippines 0 500 1000 1500 2000 2500 Megajoules per capita per day 2
Exajoules per year Annual world energy consumption 500 400 300 200 100 0 1860 1880 1900 1920 1940 1960 1980 2000 2020 Year 3
Average annual growth rates 1991-99 Country Group Annual Growth Rates (%) Population Energy consumption North America 0.9 2.7 Latin America 1.6 4.2 Western Europe 0.5 1.2 Eastern Europe 0.4-4.1 Africa 2.9 3.1 Middle East and South Asia 2.0 5.7 South-East Asia and the Pacific 1.5 6.7 Far East 1.0 4.1 4
Exponential growth Let energy consumption be N at some time t It started as N 0 at time zero (t=0) Its growth rate is dn/dt (eg barrels of oil per year) Suppose its growth rate is always proportional to its present value Then dn/dt = constant.n = kn or dn/n = k.dt Integrate this: ln(n) = kt + constant So ln(n 0 ) = 0 +constant Hence the constant is just ln(n 0 ) 5
Ctd.. Therefore ln(n) = ln(n 0 ) + kt or ln (N/N 0 ) = kt 25 Or N = N 0 e kt k is called the growth constant N N 0 20 15 10 5 0 0 0.5 1 1.5 2 2.5 3 3.5 t 6
Ctd.. If the percent growth rate is R then each year the multiplier is (1+R/100) and so N = N 0 (1+R/100) t But we have already N = N 0 e kt So k = ln (1+R/100) See example 2-1 in text 7
More doubling time How long to double the consumption if the growth constant is k? We call this the doubling time T D Set N = 2N 0 and so 2N 0 = N 0 e kt D Therefore e kt D = 2 or kt D = ln2 = 0.693 or T D = 0.693/k (sometimes called T 2 ) 8
Semilog graphs ln N = ln N o + kt is same as y = const + gradient.x ie it is a rising straight line. Plot N versus t on on semilog paper and you get a straight line its slope tells you the k value Rn N Rn N 0 0 slope = k t 9
Exajoules per year Semi-log plot of world energy consumption 1000 100 10 1 1860 1880 1900 1920 1940 1960 1980 2000 2020 Year This is a pretty good fit to a straight line with k = 0.03 yr -1 10
How long will a resource last (eg oil)? There are 2 extreme models: Constant production model (stupid model) Exponential growth model (unrealistic model in the long run) [Later we will look at a more sophisticated model due to M.K. Hubbert] Distinguish between RESERVES (known stuff, price OK) and RESOURCES (includes inferred and expected stuff, with some optimism about technology and price) 11
Constant production model This model just divides the resource by the present annual use to deduce the lifetime, ie it assumes there is no increase in use simple but wrong gives the most optimistic scenario Look at some examples : A signifies reserves B signifies resources 12
Fossil fuel reserves and resources (EJ) Region Oil reserves Oil resource N-gas reserves N-gas resource Coal reserves Coal resource World 7500 12000 6000 8000 20000 150000 Canada 1000 1200 60 4000 140 2700 USA 170 600 180 700 5000 36000 2005 production EJ/y Region Oil N-gas Coal World 164 105 122 Canada 6.1 7.0 1.5 USA 13 20 24 13
Fossil fuel lifetimes (in years) assuming constant production at 2005 level and based on Resource estimates Oil N-gas Coal World 75 75 1200 Canada 200 55 1800 USA 45 35 1500 14
See textbook re: New fossil fuels 1 Tar sands and associated limestone deposits Oil shales Hydrates CANADIAN TAR SANDS - Bitumen is extracted from sandstone and dirt mostly by surface mining. 90% of deposits need in situ extraction eg drilling wells and injecting steam under pressure. Extraction has moved in 2 decades from a risky new technology to a major oil supply industry. Future direction includes adding solvents to the steam to decrease energy use and increase output. Huge energy and water inputs are needed; there are very large greenhouse gas emissions. 15
New fossil fuels 2 LIMESTONE DEPOSITS can t be mined and they dissolve in water. Very little progress to date. New interest in electrical heating methods in 2006. COLORADO OIL SHALES are hard rock containing more than 8 trillion barrels of oil. US gov t owns 70% of formations. Shell can produce fluid oil by deep-underground heating at 370 o C for 6 months: process is energyintensive. Chevron-LANL have gov t contract starting Sept 2006 to develop green chemical methods. US DOE 2004 report states that industry could start in 2010 and by 2020 would replace 35% of US crude oil imports 16
Liquefied natural gas (LNG) Despite expected export of northern Canadian resource, US expects supply-demand gap Several LNG terminals on ocean shore under construction or planning: new pipelines will radiate from them, especially in NE USA A US market share of 1% in 2002 could be 10% in 2020 By 2020 Europe will have to import 80% of its natural gas: reliance on Russia is troubling: Germany gets 30% of its energy from Russia, expects 70% by 2025 Hence European discussion of move to LNG imports from Qatar, Trinidad, Nigeria, Iran, Indonesia 17
Extensive resource data - see BPAmoco website Many useful charts one example follows www.bp.com/centres/energy/index.asp The corresponding numerical data can be downloaded into an Excel spreadsheet. Regional data are broken down by country. There are charts and data for oil, gas, coal, nuclear, hydro and total energy. This appears to be an excellent resource but like all such data it should be treated cautiously. 18
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Exponential growth model Does not seem realistic to expect continued exp growth right up to a complete crash - but it does show a worst case scenario Consider a resource that has amount Q T remaining now - set time now as t = 0 Set lifetime as T Then Q T will be equal to integrated area of production curve from t=0 to t=t 20
Annual Production 25 20 Exponential growth model 15 10 5 N (t ) )t 0 0 0 0.5 1 1.5 2 2.5 3 3.5 time 21
Annual Production 25 20 15 10 5 0 0 Area = Q T time 0 0.5 1 1.5 2 2.5 3 3.5 T 22
Integrate the entire curve from 0 to T Q T = 0 T N0 e kt dt Q T = [ (N 0 /k) e kt ] T 0 = (N 0 /k) (e kt 1) Re-arrange: e kt = (kq T /N 0 ) + 1 And so kt = ln [ (kq T /N 0 ) + 1] Finally T = (1/k) ln [ (kq T /N 0 ) + 1] 23
T = (1/k) ln [ (kq T /N 0 ) + 1] Given the remaining resource Q T and the present production rate N 0, this equation provides the lifetime of the resource on the assumption of exponential growth right to the crash 24
Resource lifetimes with exponential growth 0% 2% 5% 10% Actual 1995-2005 World oil 75 45 31 22 1.7% Canada oil 200 80 48 31 2.6% USA oil 45 33 24 18-2.1% World gas 75 46 32 22 2.6% Canada gas 55 38 27 10 1.6% USA gas 35 27 20 15-0.2% World coal 1200 160 84 50 2.5% Can coal 1800 180 92 54-1.5% USA coal 1500 170 88 52 0.3% 25
Ctd. The above formalism was applied to our total RESOURCE estimates (see text). Try applying it yourself to the reserve data found in the BPAmoco website. And take a look at the data for other fossil fuels. 26