1/15/ Copyright 2005 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved.

Transcription

1 FIGURE 1-1 Gravitational attraction of two masses. f = k m 1m 2 f = the force attracting the two masses r 2 r = radius between centers of mass m 1, m 2 = masses k = universal gravitational constants = 6.67 x N m 2 /kg 2 1 FIGURE 1-2 Gravitational attraction of the earth. Let m1 be the mass of earth, m2 is the mass at the surface of the earth. The separation distance r is approximately the radius of the earth (if the earth is considered to be a point mass. f = m 2 g, where the gravitational constant of g = 9.81 N/kg = 9.81 m/sec 2 f and g are Vectors magnitude and direction 2 1

2 Energy Potential Energy Kinetic Energy Energy Stored in Electric and Magnetic Fields Electric Field Energy Magnetic Field Energy Nuclear Energy Table 1-1 Some Energy Equivalents Energy Unit Equivalent 1 Btu * 1 match tip 1 million Btu * 90 lb of coal * 8 gallons of propane 1 quad (10 15 Btu) * 45 million tons of coal * 10 9 ft 3 of natural gas * 170 million barrel of oil 1000 kwh (1 Mwh) of electricity * barrels of crude oil * 310 lb of coal * 3300 cu ft of natural gas 3 FIGURE 1-3 Potential energy of a mass at an elevation. Move the mass up the ramp to an altitude h, by traveling a distance l Neglecting friction between the mass and the ramp The energy required to move the mass w = mgh = mg(l sinθ) = PE (potential energy) 4 2

3 Example 1-1 An elevator in the Sears Tower in Chicago is required to lift a load of 4000 kg to an altitude of 300 m. How much energy must the motor provide (neglecting any losses in the hoist mechanism)? Solution W = 4000 kg 9.81N/kg 300m = 11,772,00 N m = MJ 1 N m = 1 Joule 1 kwh = 3.6 MJ Source: OTIS, 5 FIGURE 1-4 Pumped storage of energy to generate electricity. a. Water drives the generator during the day. b. Electricity is used to pump the water to a higher elevation during the night. 6 3

4 Example 1-2 How much energy could be stored in a pumped-storage facility if the reservoirs are separated by 400 m in altitude and each is 1.0 km square and 5.0 m deep? (Note: 1.0 cc of water has a mass of 1.0 g and 1.0 m 3 of water has a mass of 1000 kg)? Solution Amount of water (mass) to be pumped: mass = 5m x 1.0 km x 1.0 km = 5 (10 6 ) m 3 = 5(10 9 ) kg Energy needed: W = mass g h = 5x10 9 kg 9.81m/s 2 400m = x J = Tera Joule Equivalent electric energy storage:1 kwh = 3.6 MJ W = x / 3.6 x10 6 = 5.46 x 10 6 kwh =5460 Mwh 7 FIGURE 1-5 Example of storing energy to generate electricity by storing compressed air in an unused salt mine. 8 4

5 Compressed Air Energy Storage Systems Energy Storage System (EES) project Projects, Sandia National Laboratories, Lessons from Iowa: Development of a 270 Megawatt Compressed Air Energy Storage Project in Midwest Independent System Operator, January 2012, by Robert H. Schulte, Nicholas Critelli, Jr., Kent Holst amd Georgianne Huff, Sandia Report SAND , 9 FIGURE 1-6 Translational and rotational motion. Kinetic Energy: energy stored in moving mass E = 1 2 mv2 E = 1 2 Iω2 10 5

6 US DOT Connect Vehicle Technology Research V2V Safety Applications: USDOT Connected Vehicle Technology, Safety Pilot Project (V2V-SP) USDOT National Highway Traffic Safety Administration (NHTSA), DOT HS , October 2011, 0Publications/2011/ pdf Communication Infrastructures: Vehicle-to-Vehicle (V2V), Vehicle to Infrastructure, (V2I), Vehicle-to-Consumer Devices (V2D) Collected and Exchanged Data: Vehicle s latitude, longitude, time, heading angle, speed, lateral acceleration, longitudinal acceleration, yaw rate, throttle position, brake status, steering angle, headlight status, turn signal, status, vehicle length, vehicle width, vehicle mass, bumper height, and the number of occupants in the vehicle 11 Example 1-3 A 345,000 kva (345 MVA) synchronous machine is used for power factor correction and has a rotor that is 2m in diameter and 8 meter long. Assume the rotor is solid steel, with a density of 7.65 g/cm 3, and calculate the kinetic energy of the rotor is the machine runs at 600 RPM. Solution 600 RPM => 10 RPS => Angular velocity = 2π(10) rad/s = rad/s I (moment of inertia) = ½ m r 2 = ½(1932,000kg)(1m 2 ) = 96,150 kg m 2 Mass = ρv V = π r 2 l = π 1m 2 (8m) = m 3 = (10 6 ) cm 3 Mass = (10 6 ) cm 3 x (7.65 g/cm 3 ) = (10 6 )g = 192,300 kg Energy E = 1/2Iω 2 = ½(96,150 kg m 2 )(62.83 rad/s) 2 = MJ 12 6

7 FIGURE 1-7 Parallel-plate capacitor and its electric field. Capacitance C = εa d Stored energy W elec = 1/2CV 2 Farads 13 FIGURE 1-8 a. A loosely wound coil and the magnetic field resulting from the coil current. b. A tightly wound, toroidal coil in which the magnetic field is confined to the interior of the coil. Inductance L = μ 0 N 2 A 2πr Energy stored in an inductor W mag = ½ L I

8 FIGURE 1-9 Components of the power system. 15 FIGURE 1-13 Conceptual diagram of a coal-fired power plant. 16 8

9 FIGURE 1-14 U.S. generation capacity during the twentieth century. 17 FIGURE 1-16 Simple power system. 18 9

10 FIGURE 1-17 Simplified one-line diagram of a power system. 19 FIGURE 1-18 Typical variation of electrical load during a week

11 FIGURE 1-19 Graph of typical electrical frequency fluctuations. 21 FIGURE 1-20 Map showing the North American Electric Reliability Council regions

12 FIGURE 1-21 Number of utility nuclear generating plants over time. 23 FIGURE 1-22 Status of electric utility restructuring in the United States. (Source: U.S. Dept. of Energy.) 24 12

13 FIGURE 1-23 Electrical generation by utilities and nonutilities. (Source: U.S. Dept. of Energy, March 2003, Monthly Energy Update.) 25 FIGURE 1-24 Annual, per-capita greenhouse gas emissions for a number of industrialized nations

14 FIGURE 1-25 Change of population, per-capita emissions, GDP, and emissions per dollar of GDP from 1990 to FIGURE 1-26 Monthly average cost per MWhr in California

15 FIGURE 1-27 Price of Enron stock from 1992 to FIGURE 1-28 System for Problem