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Energy, Environment, and Industrial Development. Michael B. McElroy Frederick H. Abernathy Lecture 16 April 10, 2006. An atom in its normal state is electrically neutral. If it loses an electron, it assumes a positive charge and is known as a positive ion.
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Energy, Environment, and Industrial Development Michael B. McElroy Frederick H. Abernathy Lecture 16 April 10, 2006
An atom in its normal state is electrically neutral. If it loses an electron, it assumes a positive charge and is known as a positive ion. • The fundamental unit of negative charge is that carried by an electron
Charge in the SI system of units is expressed in units of coulombs (C): French physicist Charles-Augustin de Coulomb (1736-1806). The charge on an electron = -1.602x10-19 C. • How many electrons do you need to provide a charge of -1 C?
Consider the electrostatic force between particles of charge q1 and q2. The force on q1 due to the presence of q2 is given by • This is known as Coulomb’s Law q1 r q2
Suppose the charges q1 and q2 have opposite signs. For example, suppose q2 represents the charge on a proton and q1 the charge on an electron as in a hydrogen atom. • The force is now directed opposite to : it acts to attract particles of opposite sign. • Particles of the same sign are repelled
Coulomb’s Law: • With charges expressed in units of Coulombs (C) and distance in m, F is in Newtons (N), with k = 8.99x109Nm2C-2 • Check units: N Nm2C-2 C2 m-2
Example A1.14 • The charge passing position P in the conductor in unit time defines what is known as current • If charge Δq passes P in time Δt, then I = Δq/ Δt defines current • I has dimensions of charge per unit time, Coulomb sec-1 • The unit of current is the ampere (A) honoring Andre-Marie Ampere (1775-1836) P
The electrostatic force on a particle of charge q is given by qE • Here E is a vector known as the electric field • The gravitational force on a particle of mass m is given by mg, where g is the acceleration of gravity • E, the electric field, is analogous to the field defining the gravitational force experienced by a particle of unit mass
To move a particle of charge q through a displacement Δr in the presence of an electric field E requires an input of work • If we wish to move q in a direction opposite to E, then is negative. Hence ΔW is positive. Work must be done to move a (positive) charge q against the direction of the electric field. • Work must be done to move a mass m up against the gravitational field.
Work done to move unit charge from ab in the presence of an electric field E: • V is known as the electric potential or simply as the potential. The potential is expressed in units of Volts (V)
It follows that the electric field has dimensions of Vm-1 • A positive charge placed in the electric field E will accelerate in the direction of the field: ΔW < 0 Vb – Va < 0 Vb < Va • The motion proceeds from high to low voltage • Gravitational analogue: If mass falls from ab, its kinetic energy increases, its potential energy decreases
A material with the property that it can maintain a net flow of charge is known as a conductor. Examples: copper or aluminum wire. • In the presence of an electric field, or equivalently a voltage differential, electrons will move • Electrons move from low to high voltage: current flows from high to low as though charge was transferred by positively charged particles. • 1A is equivalent to a flow of charge equal to 1C sec-1
Ohm’s Law, named for Bavarian George Simon Ohm (1789-1854) defines a relation between current and voltage: ΔV = RI • R is known as the resistance. R has dimensions of V A-1 • The unit of R in the SI system is know as the ohm (Ω)
For a wire of length L and cross section A, R = r L/A where r, known as the resistivity, is a property of the medium • r has dimensions of ohm meters
The loss of electrical energy per unit time due to movement of charge from ab through a voltage drop V is given by multiplying the charge transferred per unit time by the work exerted by the electric field on unit charge • Using Ohm’s Law P = IV = I (RI) = I2R • Or, P = I2 r (L/A)
To maintain a steady current I a conductor requires a continuous input of energy. This is referred to as a seat of electromotive force or simply a source of emf. • The seat of emf maintains the voltage differential required to drive the current
Charged particles experienced a force due not only to the electric field but also due to the magnetic field F = q v x B • With F in N, q in C, v in m/s, B has dimensions of NC-1m-1s or N A-1 m-1 • The unit of magnetic field in the SI system is the tesla (T) – Serbian-American Nikola Tesla (1856-1943). • Strength of the Earth’s magnetic field at mid latitudes is about 7x10-5T = 0.7 Gauss (G)
A current can produce a magnetic field • Intensity of the magnetic field defined by the Biot-Savart Law. • To find the direction of the magnetic field at pt. P, place your thumb along direction of current flow at Q extend hand towards P curl of fingers with indicate direction of B
For a current flowing in a long straight wire Figure A1.11
Consider currents flowing in 2 contiguous wires, 1 and 2. Assume wires are long, straight, and parallel • The force on a length l of 2 due to wire 1 is given by Here R defines the separation of the wires • If the currents are flowing in the same direction, the wires are drawn together. If currents are flowing in opposite directions, wires are driven apart
Ampere’s Law allows for an alternative way to calculate the strength of the magnetic field produced by a current Figure A1.12
For a circular path Figure A1.12
A strong magnetic field can be formed inside the solenoid where n is the number of loops of wire per unit length of the solenoid Figure A1.12
Concept of magnetic flux Фm = B.n ∆A • If B is constant over the area and perpendicular to the area, then Фm = B A Figure A1.14
Faraday’s Law, named for English physicist Michael Faraday (1791-1841) states that Electromotive force Figure A1.14
Consider coil rotating at a uniform rate ω θ= ωt • At orientation θ, Фm = BAcosωt • ε(t) = BAωsinωt • ε oscillates in time • Since, by Ohm’s Law, ε = IR • Example of an alternating current
If number of turns in secondary circuit is larger than in primary, voltage is increased. If smaller, voltage is decreased. Step-up or step-down transformer
Development of the US electric power system • Beginning of modern electric industry, 1882 • Edison’s Pearl Street generating station operational on Sep. 4, 1882 • Consumed 10 pounds of coal per kilowatt-hour • Served 59 customers charging 24 cents/ kilowatt-hour • By end of 1880’s small central stations in many US cities • Development of hydroelectric plant at Niagara Falls by George Westinghouse in 1896. Delivered power to Buffalo, 20 miles away
Development of the US electric power system • Municipally owned utilities supplied street lighting and trolley services. Accounted for 8% of total power generation in 1900 • Residential rate fall to <17 cents a kilowatt-hour • Consolidation in generating industry. By late 1920s, 16 companies controlled >75% of total US generating capacity • State regulation of utilities. Later federal involvement with creation of Federal Power Commission in 1920 • Electric power capacity grew at ~12% per year from 1901-1932
Development of the US electric power system • Electricity prices dropped to 5.6 cents per kilowatt-hour in 1932 • By 1932, 67% of residences supplied with electricity –80% of urban dwellings. But, only 11% of farms had electricity • Rural Electrification Act of 1936 established the Rural Electrification Administration • By 1941, 35% of farms were electrified. • Hoover Dam, 1936; Grand Coulee 1941 • Electricity prices in 1941, 3.73 cents a kilowatt-hour. Half of all farms electrified by 1945
Development of the US electric power system • From 1945-1950, electricity use grew at >8% per year. Prices continued to decline. 80% of farms electrified by 1950 • Generation increased by >8.5% per year from 1950-1960. Commercial nuclear power introduced. • During 1960’s environmental concerns with power generation begin to have influence