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
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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.
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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)
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distance east
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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
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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
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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)
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i
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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
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t  0
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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
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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
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 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
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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
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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)
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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
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3 Pa
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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
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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
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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
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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
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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
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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
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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
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US energy consumption (1998)
a: 18.92 million barrels / day (cost, $851 million/day)
b: 21.34 tcf/yr
Source:
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http://energy.cr.usgs.gov/e
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nergy/stats_ctry/stat1.html
Consumption of Energy by Sector for US
1998
Source: http://energy.cr.usgs.gov/energy/stats_ctry/stat1.html
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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%
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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
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US Energy Consumption by Sector (1973-2000)
transportation
industrial
commercial
residential
120000
trillion BTU
100000
80000
60000
40000
20000
Source: Energy Information Administration,
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Monthly Energy ReviewSpring
December
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1999
1997
1995
1993
1991
Year
1989
1987
1985
1983
1981
1979
1977
1975
1973
0
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US Energy Consumption by Source (1973-2000)
120000
Trillion BTU
100000
Petroleum Products
Natural Gas
Coal
Nuclear Electric Power
Renewables
80000
60000
40000
20000
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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,
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Monthly Energy ReviewSpring
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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
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States
2003
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Source: Energy Information Administration
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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.
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

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
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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.
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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.
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