Transcript aabvzvzcx - Harvard University

Energy, Environment, and
Industrial Development
Michael B. McElroy
Frederick H. Abernathy
Lecture 16
April 10, 2006
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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
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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 = -1.602x1019 C.
How many electrons do you need to
provide a charge of -1 C?
1
1019
18

 6.2 x10
19
1.602
1.602x10
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Consider the electrostatic force
between particles of charge q1 and
q2. The force on q1 due to the
rˆ
presence of q2 is given by
Fq1  kq 1 q 2
r
q2

r
2
q1
r rˆ  r
This is known as Coulomb’s Law
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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.
rˆ
Fq1  k q 1 q 2


r
2
The force is now directed opposite to
rˆ : it acts to attract particles of opposite
sign.
Particles of the same sign are repelled
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Coulomb’s Law:
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With charges expressed in units of
Coulombs (C) and distance in m, F
is in Newtons (N), with k =
8.99x109Nm2C-2
Check units:

F
N

rˆ
F  kq 1 q 2 2
r
k
Nm2C-2
q1 q 2
C2
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rˆ
2
r
m-2
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Example A1.14
The charge passing position P in the
conductor in unit time defines what is
known as current
P
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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 (17751836)
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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
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To move a particle of charge q through a
displacement Δr in the presence of an
electric field E requires an input of work
W  q E  r
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If we wish to move q in a direction
opposite to E, then E  r 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.
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Work done to move unit charge from
ab in the presence of an electric
field E:
b
W  Vb  Va    E  r
a
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V is known as the electric potential
or simply as the potential. The
potential is expressed in units of
Volts (V)
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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
ab, its kinetic energy increases, its
potential energy decreases
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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
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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)
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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
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The loss of electrical energy per unit time
due to movement of charge from ab
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)
q
P(
)V  IV
t
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Figure A1.8
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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
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Figure A1.9
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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) – SerbianAmerican Nikola Tesla (1856-1943).
Strength of the Earth’s magnetic field at
mid latitudes is about 7x10-5T = 0.7
Gauss (G)
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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
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Figure A1.10
l  rˆ
B  k m I
r2
7
2
k m  10 NA
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For a current flowing in a long
straight wire
I
B  2k m I
R
Figure A1.11
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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 F  2k l I I
R
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
1
2
2
m
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Ampere’s Law allows for an
alternative way to calculate the
strength of the magnetic field
produced by a current
 B  l  4k
m
I
Figure A1.12
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 B  l  4k I
For a circular path
m
2RB  4k m I
B
2k m I
R
Figure A1.12
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Figure A1.12
A strong magnetic field can be formed inside
the solenoid
B  4k m nI
where n is the number of loops of wire per unit
length of the solenoid
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Concept of magnetic flux Фm = B.n ∆A
Figure A1.14
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If B is constant over the area and
perpendicular to the area, then Фm = B A
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d m
dt
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Faraday’s Law, named for English
physicist Michael Faraday (1791-1841)
states that    d m
dt
Electromotive force
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Figure
A1.14
d m
dt
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Consider coil rotating at a uniform rate
ω  θ= ωt
At orientation θ, Фm = BAcosωt
d m
  BA sin t
dt
ε(t) = BAωsinωt
ε oscillates in time
Since, by Ohm’s Law, ε = IR 
1
BA 
I (t )   (t ) 
sin t
R
R
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Example of an alternating current
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d m
dt
B 2 A2 2
P(t ) I (t ) R  
 sin 2 t
R
1 B 2 A 2 2
Pav 
2
R
2
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Figure A1.16
d m
dt
d m
 N 2
dt
V
 N2 ( 1 )
N1
V1   N 1
V2
V2
N2
V2  (
)V1
N1
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d m
dt
V2  (
N2
)V1
N1
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
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Development of the US electric power system
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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
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Development of the US electric power system
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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 kilowatthour
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
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from 1901-1932
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Development of the US electric power system
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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 kilowatthour. Half of all farms electrified by 1945
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Development of the US electric power system
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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
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