#### Transcript aabvzvzcx - Harvard University

Energy, Environment, and Industrial Development Frederick H. Abernathy Michael B. McElroy Lecture 3 Feb. 8, 2006 Basic Physical Quantities Length: SI unit; meter (m) British/American units; inches, feet, miles Mass: SI unit; kilogram (kg) British/American units; ounces, pounds, tons Time: SI unit; second (s) British/American units; minutes, hours, years 2 Spring [email protected] + MBM Science A-52 Basic Physical Quantities Scalar quantities are completely specified by a number and appropriate unit: 500 kg 3x107 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. 3 Spring [email protected] + MBM Science A-52 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 distance north P(x,y) o origin (Boston) x Spring [email protected] + MBM Science A-52 distance east 4 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) j P r yj θ xi i 5 Spring [email protected] + MBM Science A-52 Definition of vector dot product A · B = |A| |B| cosθ where θ is the angle between A and B j A θ A · B is a scalar. B i Velocity is a vector r · v would have dimension of r x v m x (m s-1) = m2 s-1 6 Spring [email protected] + MBM Science A-52 Displacement Vector 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. j r (t) θ Δr r (t+Δt) r (t) + Δ r = r (t+Δt) Δ r = r (t+Δt) - r (t) Spring [email protected] + MBM Science A-52 i 7 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. r d r Instantaneous v(t) = lim t dt Velocity has units of m s-1 distance traveled per unit time 8 t 0 Spring [email protected] + MBM Science A-52 Acceleration defines the change in velocity per unit time v(t t ) v(t ) a t average acceleration over Δt v(t t ) v(t ) d v a lim t 0 t dt instantaneous acceleration over Δt Acceleration is a vector: it defines both the rate of change of velocity but also the direction of change 9 Spring [email protected] + MBM Science A-52 Consider a mass m moving along the circumference of a circle at constant speed v Figure taken from McElroy 2001 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 10 Spring [email protected] + MBM an accelerationScience A-52 Definition of linear momentum: p = mv Newton’s law of motion: Force = rate of change of linear momentum If mass is constant dp dv m ma F ma dt dt 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 11 Spring [email protected] + MBM Science A-52 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 12 Spring [email protected] + MBM Science A-52 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 1m2 of surface contains 1.035x104 kg m-2. To calculate the weight of the atmosphere, or the force exerted on the surface, multiply by g. P = (1.035x104 kg m-2) (9.8 ms-2) = 1.014 x105 kg m-1s-2 (or N m-2) Unit of pressure in SI system is the Pascal (Pa) Patm = 1.014 x105 Pa = 101.4 kilopascal (kPa) (actually 101.3 kPa) 13 Spring [email protected] + MBM Science A-52 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 [email protected] + MBM 3 Pa 1 psi = 6.895x10Spring Science A-52 14 Power is a measure of the rate at which work is done by a force F r r P ,v t t P F v Power has units of J s-1 or kg m2 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 15 Spring [email protected] + MBM Science A-52 The SI (système International) system of Units 1. Length – meter (m) Mass – kilogram (kg) Time – second (s) 2. Force – Newton (N) 3. Energy – Joule (J) 4. Kinetic energy = ½ mv2 Mass of 1kg moving at 1m/s has kinetic energy of 1 J 16 Spring [email protected] + MBM Science A-52 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 = 106 g Volume: 1 gallon (UK) = 4.54609 liters 1 gallon (US) = 3.785412 liters 1 liter = 103 cm3 = 10 -3 m3 1000 liters = 1 m3 17 Spring [email protected] + MBM Science A-52 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) = 103 cal 1 BTU (British Thermal Unit) is the heat required to raise the temperature of 1 pound of water by 1°F 1 Quad = 1015 BTU 1 Therm = 105 BTU 1 BTU = 1055 J = 1.055 x 103 J 18 Spring [email protected] + MBM Science A-52 Other Units 1 Barrel of oil = 42 gallons = 159 liters Energy content of 1 barrel of oil = 6.12 x 109 J = 1700 KW hr 1 BTU = 1.055 x 103 J Energy content of 1 barrel of oil = ( 6.12 x 109 / 1.055 x 103 ) BTU = 5.8 x 106 BTU 1 Quad = 1015 BTU = (1015 / 5.8 x 106 ) barrels of oil = 1.7 x 108 barrels 19 Spring [email protected] + MBM Science A-52 More Cost / barrel = $45 Cost / quad oil = $ 1.7 x 108 x 4.5 x 101 = $7.65 x 109 Total energy consumption in the US ~ 95 Quad $7.65 x 109 x 9.5 x 101 = $ 7.3 x 1011 = $ 730 billion Actual US oil consumption ~ 37 Quad $7.65 x 109 x 3.7 x 101 = $ 280 billion Spring [email protected] + MBM Science A-52 20 Energy content of fuels Coal Crude oil Oil Gasoline Natural gas Wood 25 million BTU / ton 5.6 million BTU/barrel 5.78 million BTU/barrel = 1700 kWh 5.6 million BTU/barrel 1030 BTU/cubic foot 20 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 21 Spring [email protected] + MBM Science A-52 US energy consumption (1998) a: 18.92 million barrels / day (cost, $851 million/day) b: 21.34 tcf/yr Source: 22 http://energy.cr.usgs.gov/e c: 1038 million short tonsSpring /yr [email protected] + MBM Science A-52 nergy/stats_ctry/stat1.html Consumption of Energy by Sector for US 1998 Source: http://energy.cr.usgs.gov/energy/stats_ctry/stat1.html 23 Spring [email protected] + MBM Science A-52 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% Spring [email protected] + MBM Science A-52 24 25 Spring [email protected] + MBM Science A-52 http://eed.llnl.gov/flow/02flow.php Total number of US households: 107 x 106 Energy consumption per household member: 35.9 x 106 BTU Energy consumption per household member for houses built before: 1939 46 x 106 BTU 1940 -1949 37.5 x 106 BTU 1950 – 1959 38.9 x 106 BTU 1960 – 1969 35.6 x 106 BTU 1970 – 1979 31.4 x 106 BTU 1980 – 1989 31.7 x 106 BTU 1990 – 1999 31.3 x 106 BTU 2000 – 2001 33.1 x 106 BTU 26 Spring [email protected] + MBM Science A-52 US Energy Consumption by Sector (1973-2000) transportation industrial commercial residential 120000 trillion BTU 100000 80000 60000 40000 20000 Source: Energy Information Administration, [email protected] + 2004 MBM Monthly Energy ReviewSpring December Science A-52 1999 1997 1995 1993 1991 Year 1989 1987 1985 1983 1981 1979 1977 1975 1973 0 27 US Energy Consumption by Source (1973-2000) 120000 Trillion BTU 100000 Petroleum Products Natural Gas Coal Nuclear Electric Power Renewables 80000 60000 40000 20000 19 73 19 75 19 77 19 79 19 81 19 83 19 85 19 87 19 89 19 91 19 93 19 95 19 97 19 99 0 Year Source: Energy Information Administration, [email protected] + 2004 MBM Monthly Energy ReviewSpring December Science A-52 28 US Carbon Dioxide Emissions from Energy Consumption by End-Use sector, 1990 and 19952003 Transportation 7000 Million Metric Tons Carbon Dioxide 6000 Industrial Commercial Residential 5000 4000 3000 2000 1000 0 1990 1995 1996 1997 1998 1999 2000 2001 2002 2003 Year In 2003, electricity generation accounts for ~39% of CO2 emissions Source: Energy Information Administration, Emissions of Spring [email protected] + MBM Greenhouse Gases in the United States 2003 Science A-52 29 Source: Energy Information Administration Spring [email protected] + MBM Science A-52 30 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. 31 Spring [email protected] + MBM Science A-52 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)(1.81x102m) = 1.77x103 kg m2s-2 = 1.77 x103 J ½ mv2 = 1.77 x 103 kg m2s-2 m = 1 kg v = 57.8 ms-1 32 Spring [email protected] + MBM Science A-52 Example 2 The flow of water in the Yangtze River immediately upstream of he dam averages 60,000 m3 s-1. Of this, 20,000 m3 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. 33 Spring [email protected] + MBM Science A-52 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 103 J. The mass corresponding to 1 m3 of water = 103 kg. The power generated = (kg s-1 of water flow) x (J kg-1) = (2x107 kg s-1) x (1.22 x 103 J kg-1) = 2.44 x 1010 J s-1 = 24.4 x 109 W The dam has the potential to generate 24.4 gigawatts (GW) of electrical energy (a gigawatt = 109w). 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. 34 Spring [email protected] + MBM Science A-52