1. Energy, Environment, and Industrial Development Frederick H. Abernathy Michael B. McElroy Lecture 3 Feb. 8, 2006

2. Spring 2006@FHA + MBM Science A-52 2 Basic Physical Quantities Length: SI unit; meter (m) British/American units; inches, feet, miles Mass: SI unit; kilogram (kg) SI unit; kilogram (kg) British/American units; ounces, pounds, tons British/American units; ounces, pounds, tonsTime: SI unit; second (s) SI unit; second (s) British/American units; minutes, hours, years British/American units; minutes, hours, years

3. Spring 2006@FHA + MBM Science A-52 3 Basic Physical Quantities Scalar quantities are completely specified by a number and appropriate unit: 500 kg 3x10 7 sec 500 m Vector quantities are specified by a number, with appropriate unit, and a direction. At a given moment in time, a person was located 100 miles from Boston in a direction to the north east.

4. Spring 2006@FHA + MBM Science A-52 4 Basic Physical Quantities Example of a vector: the position vector, written as r Can identify r using a coordinate system with a suitable origin (Boston) r can be defined using 2 numbers (distance east, distance north) y x o origin (Boston) distance north distance east P(x,y)

5. Spring 2006@FHA + MBM Science A-52 5 j i P r x ix i y jy j Alternatively: Alternatively: r = x i + y j Means  move a distance x in i direction (east) and then move a distance y in j direction (north) θ

6. Spring 2006@FHA + MBM Science A-52 6 j i Definition of vector dot product A · B = |A| |B| cosθ where θ is the angle between A and B A · B is a scalar. Velocity is a vector r · v would have dimension of r x v  m x (m s -1 ) = m 2 s -1 θ A B

7. Spring 2006@FHA + MBM Science A-52 7 Displacement Vector j i Suppose an object is located at Suppose an object is located at r at time t. Write as r (t) Suppose a little later, at time t + Δt, it is at r (t+Δt). Difference between r (t+Δt) and r (t) defines the displacement vector. θ r (t+Δt) r (t) ΔrΔr r (t) + Δ r = r (t+Δt) Δ r = r (t+Δt) - r (t)

8. Spring 2006@FHA + MBM Science A-52 8 Velocity, a vector, is a measure of distance traveled per unit time, including information on direction. Displacement between t, t+Δt = Δr Time interval = Δt Velocity = v = Δr / Δt Strictly speaking this defines the average velocity for time interval Δt. Instantaneous v(t) = Velocity has units of m s -1  distance traveled per unit time

9. Spring 2006@FHA + MBM Science A-52 9 Acceleration defines the change in velocity per unit time Acceleration is a vector: it defines both the rate of change of velocity but also the direction of change Δt average acceleration over Δt Δt instantaneous acceleration over Δt

10. Spring 2006@FHA + MBM Science A-52 10 Consider a mass m moving along the circumference of a circle at constant speed v Between time t and t + Δt, v has turned through an angle ΔФ. There is no change in v, but there is a change in v  an acceleration Figure taken from McElroy 2001

11. Spring 2006@FHA + MBM Science A-52 11 Definition of linear momentum: p = mv Newton’s law of motion: Force = rate of change of linear momentum If mass is constant  Force is a vector. Unit force (magnitude) corresponds to a mass of 1 kg accelerating at a rate of 1 m/s per sec = 1m s -2 Unit force is the Newton

12. Spring 2006@FHA + MBM Science A-52 12 Definition of Work If under the action of a force F, a mass is displaced through a displacement vector Δr, we say that the force has done work on the mass m, the quantity of which is given by ΔW = F · Δr Suppose the mass is moved in the direction of the force, then ΔW = F |Δr| Work is a scalar. If the object moves in a direction opposite to the applied force, ΔW is negative

13. Spring 2006@FHA + MBM Science A-52 13 Pressure is a measure of force per unit area: Nm -2 The atmosphere exerts pressure: the entire mass of the atmosphere is being carried by the surface. The atmosphere above 1m 2 of surface contains 1.035x10 4 kg m -2. To calculate the weight of the atmosphere, or the force exerted on the surface, multiply by g. P = (1.035x10 4 kg m -2 ) (9.8 ms -2 ) = 1.014 x10 5 kg m -1 s -2 (or N m -2 ) = 1.014 x10 5 kg m -1 s -2 (or N m -2 ) Unit of pressure in SI system is the Pascal (Pa) P atm = 1.014 x10 5 Pa = 101.4 kilopascal (kPa) (actually 101.3 kPa) (actually 101.3 kPa)

14. Spring 2006@FHA + MBM Science A-52 14 Pressure of atmosphere at surface is approximately the same as the pressure exerted by a column of water 10m high. Pressure can be measured by a barometer. Height of the mercury column  0.76m Pressure of the atmosphere  14.7 psi 1 psi = 6.895x10 3 Pa

15. Spring 2006@FHA + MBM Science A-52 15 Power is a measure of the rate at which work is done by a force Power has units of J s -1 or kg m 2 s -3. Unit of power in mks (SI) system  Watt (W) 1 Watt corresponds to expenditure of work at a rate of 1 J s -1

16. Spring 2006@FHA + MBM Science A-52 16 The SI (système International) system of Units 1. Length – meter (m) Mass – kilogram (kg) Mass – kilogram (kg) Time – second (s) Time – second (s) 2. Force – Newton (N) 3. Energy – Joule (J) 4. Kinetic energy = ½ mv 2 Mass of 1kg moving at 1m/s has kinetic energy of 1 J

17. Spring 2006@FHA + MBM Science A-52 17 UK (Imperial) and US units Length: inches, feet, yards, miles 1 yard = 0.9144 m Mass: ounces, pounds, stones, tons 1 pound = 0.453592 kg Differences between UK and US ton 1 ton (UK) = long ton = 2240 pounds 1 ton (US) = short ton = 2000 pounds 1 metric ton = 1000 kg = 10 6 g Volume: 1 gallon (UK) = 4.54609 liters 1 gallon (US) = 3.785412 liters 1 gallon (US) = 3.785412 liters 1 liter = 10 3 cm 3 = 10 -3 m 3 1000 liters = 1 m 3

18. Spring 2006@FHA + MBM Science A-52 18 Energy Units  1 calorie (cal) is the heat required to raise the temperature of 1 g of water by 1 degree centigrade  Food calories are actually kilocalories  1 food calorie (Cal) = 10 3 cal  1 BTU (British Thermal Unit) is the heat required to raise the temperature of 1 pound of water by 1°F  1 Quad = 10 15 BTU  1 Therm = 10 5 BTU  1 BTU = 1055 J = 1.055 x 10 3 J

19. Spring 2006@FHA + MBM Science A-52 19 Other Units 1 Barrel of oil = 42 gallons = 159 liters Energy content of 1 barrel of oil = 6.12 x 10 9 J = 1700 KW hr = 1700 KW hr 1 BTU = 1.055 x 10 3 J Energy content of 1 barrel of oil = ( 6.12 x 10 9 / 1.055 x 10 3 ) BTU = 5.8 x 10 6 BTU 1 Quad = 10 15 BTU = (10 15 / 5.8 x 10 6 ) barrels of oil = (10 15 / 5.8 x 10 6 ) barrels of oil = 1.7 x 10 8 barrels = 1.7 x 10 8 barrels

20. Spring 2006@FHA + MBM Science A-52 20 More Cost / barrel = $45 Cost / quad oil = $ 1.7 x 10 8 x 4.5 x 10 1 = $7.65 x 10 9 = $7.65 x 10 9 Total energy consumption in the US ~ 95 Quad ~ 95 Quad  $7.65 x 10 9 x 9.5 x 10 1 = $ 7.3 x 10 11 = $ 730 billion Actual US oil consumption ~ 37 Quad  $7.65 x 10 9 x 3.7 x 10 1  $7.65 x 10 9 x 3.7 x 10 1 = $ 280 billion

21. Spring 2006@FHA + MBM Science A-52 21 Energy content of fuels Coal 25 million BTU / ton Crude oil5.6 million BTU/barrel Oil5.78 million BTU/barrel = 1700 kWh Gasoline5.6 million BTU/barrel Natural gas1030 BTU/cubic foot Wood20 million BTU/cord In the US, the cord is defined legally as the volume of a stack of firewood 4 feet wide, 8 feet long, and 4 feet high. 1 cord = 128 cubic feet = 3.6247 cubic meters

22. Spring 2006@FHA + MBM Science A-52 22 US energy consumption (1998) a: 18.92 million barrels / day (cost, $851 million/day) b: 21.34 tcf/yr c: 1038 million short tons /yr Source: http://energy.cr.usgs.gov/e nergy/stats_ctry/stat1.html

23. Spring 2006@FHA + MBM Science A-52 23 Consumption of Energy by Sector for US Source: http://energy.cr.usgs.gov/energy/stats_ctry/stat1.html 1998

24. Spring 2006@FHA + MBM Science A-52 24 World energy production (1998) Source: http://energy.cr.usgs.gov/energy/stats_ctry/stat1.html US fraction = 94.27 / 379.7 = 24.8% US population fraction = 280 million / 6 billion = 4.7%

25. Spring 2006@FHA + MBM Science A-52 25 http://eed.llnl.gov/flow/02flow.php

26. Spring 2006@FHA + MBM Science A-52 26  Total number of US households: 107 x 10 6  Energy consumption per household member: 35.9 x 10 6 BTU  Energy consumption per household member for houses built before: 1939  46 x 10 6 BTU 1940 -1949  37.5 x 10 6 BTU 1950 – 1959  38.9 x 10 6 BTU 1960 – 1969  35.6 x 10 6 BTU 1970 – 1979  31.4 x 10 6 BTU 1980 – 1989  31.7 x 10 6 BTU 1990 – 1999  31.3 x 10 6 BTU 2000 – 2001  33.1 x 10 6 BTU

27. Spring 2006@FHA + MBM Science A-52 27 Source: Energy Information Administration, Monthly Energy Review December 2004

28. Spring 2006@FHA + MBM Science A-52 28 Source: Energy Information Administration, Monthly Energy Review December 2004

29. Spring 2006@FHA + MBM Science A-52 29 In 2003, electricity generation accounts for ~39% of CO 2 emissions Source: Energy Information Administration, Emissions of Greenhouse Gases in the United States 2003

30. Spring 2006@FHA + MBM Science A-52 30 Source: Energy Information Administration

31. Spring 2006@FHA + MBM Science A-52 31 Example 1 The Chinese government is nearing completion of what will eventually be the world’s largest dam, the Three Gorges Dam on the Yangtze River. The dam will extend to a height of 181 m. Assume that 1 kg of water is allowed to overflow the dam and fall to the bottom on the other side. Calculate its kinetic energy when it reaches the bottom and the speed of the water.

32. Spring 2006@FHA + MBM Science A-52 32 Answer: The kinetic energy of the water when it reaches bottom is equal to its potential energy with respect to the base of the dam at the point of overflow: Kinetic Energy = mgh = (1kg)(9.8ms -2 )(1.81x10 2 m) = (1kg)(9.8ms -2 )(1.81x10 2 m) = 1.77x10 3 kg m 2 s -2 = 1.77x10 3 kg m 2 s -2 = 1.77 x10 3 J = 1.77 x10 3 J ½ mv 2 = 1.77 x 10 3 kg m 2 s -2 ½ mv 2 = 1.77 x 10 3 kg m 2 s -2 m = 1 kg  v = 57.8 ms -1

33. Spring 2006@FHA + MBM Science A-52 33 Example 2 The flow of water in the Yangtze River immediately upstream of he dam averages 60,000 m 3 s -1. Of this, 20,000 m 3 s -1, is used to drive turbines to generate electricity. The turbines are situated 125 m below the level of the water behind the dam. We refer to this as the hydraulic head for the water driving the turbines. Estimate the electrical power that would be realized if 100% of the potential energy of the water flowing through the turbines could be converted to electricity.

34. Spring 2006@FHA + MBM Science A-52 34 Answer: Following the analysis in the previous example, the change in energy associated with 1 kg of water falling 125 m is equal to 1.22 x 10 3 J. The mass corresponding to 1 m 3 of water = 10 3 kg. The power generated = (kg s -1 of water flow) x (J kg -1 ) = (2x10 7 kg s -1 ) x (1.22 x 10 3 J kg -1 ) = (2x10 7 kg s -1 ) x (1.22 x 10 3 J kg -1 ) = 2.44 x 10 10 J s -1 = 2.44 x 10 10 J s -1 = 24.4 x 10 9 W = 24.4 x 10 9 W The dam has the potential to generate 24.4 gigawatts (GW) of electrical energy (a gigawatt = 10 9 w). The design objective is 18.2 GW, which can be realized by converting the potential energy of the water to electricity with an efficiency of about 75%, a realistic expectation for hydroelectric power generation with modern technology.