Transcript Type title of presentation here Maximum of two lines

CHAPTER 13
Energy and Power
© 2011 Cengage Learning Engineering. All Rights Reserved.
13-1
Outline
In this chapter we will
• introduce concepts of energy and power
• explain various forms of energy
• explain the term efficiency
• discuss the importance of energy sources,
generation, and consumption
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13-2
What Is Energy?
• Energy is one of those abstract terms that you
already have a good feel for. It represents the
ability to do work
• Energy can have different forms
 kinetic energy
 potential energy
 elastic energy
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13-3
Kinetic Energy
When work is done on or against an object, it changes
the kinetic energy of the object
Mechanical work done by the engine
brings about a change in kinetic energy
work1 2
1
1
2
 mV2  mV12
2
2
© 2011 Cengage Learning Engineering. All Rights Reserved.
1
1
2
KE  mV2  mV12
2
2
13-4
Kinetic Energy
Units:
N
2
1
m
m
2
kinetic energy  mV  kg    kg  2 m 
2
s
s 
 N  m  joule  J
It is important to note that it is the change in
kinetic energy (KE) that is used in engineering
analysis
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13-5
Example 13.1 – Work
Given: a car traveling at 90 km/h has a mass of 1400 kg.
Find: the net force needed to bring the car to a full stop in
a distance of 100 m.
Solution:
m
 km  1 h  1000 m 
V1  Vinitial  90


  25
s
 h  3600 s  1 km 
V2  Vfinal  0
work 1-2  force distance  
1
1
mV22  mV12
2
2
force 100 m   0  1 1400 kg 25 m/s 2
2
force  4375 N
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13-6
Potential Energy
The gravitational potential energy
represents the amount work
required to lift an object with a
mass m through a vertical distance
h. It is the work required to
overcome the gravitational pull of
the earth.
PE  mgh
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Δh
m
13-7
Potential Energy
Units:
N
m
potential energy  mgh  kg  2 m 
s 
 N  m  joule  J
It is important to note that it is the change in
potential energy (PE) that is used in
engineering analysis
© 2011 Cengage Learning Engineering. All Rights Reserved.
13-8
Example 13.2 – Potential Energy
Given: an elevator and its occupant with a
mass of 2000 kg
Find: the energy required to lift the elevator
and its occupant (a) between 1st and 2nd
floor, (b) between 3rd and 4th floor, (c)
between 1st and 4th floor.
Solution:
(a) change in potential energy = mgh


 2000 kg  9.81 m/s 2 4.5 m  88,290 J
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13-9
Example 13.2 – Potential Energy
Solution (continued):
(b) change in potential energy between third and fourth floor


 2000 kg  9.81 m/s 2 4.5 m  88,290 J
(c) change in potential energy between first and fourth floor


 2000 kg  9.81 m/s 2 13.5 m  264,870 J
Note: the amount of energy required to lift the elevator
from 1st to 2nd floor and from 3rd to 4th floor is the same.
We have neglected any frictional effect in our analysis.
The actual energy requirement would be greater in the
presence of frictional effect.
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13-10
Elastic Energy
• When a spring is stretched or
compressed from its unstretched
position, elastic energy is stored
in the spring.
• Energy will be released when the
spring is allowed to return to its
unstretched position
• Elastic energy stored in the spring
is given by:
1 2
elastic energy  kx
2
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13-11
x
Force
Force
Elastic Energy
Units:
1 2 N 2
elastic energy  kx   m 
2
m
 N  m  joule  J
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13-12
Elastic Energy
1
elastic energy  kx12
2
1
elastic energy  kx22
2
change in elastic energy  EE
x1
x2
Force
Force
Force
Force
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1 2 1 2
 kx2  kx1
2
2
13-13
Example 13.3 – Elasticity Energy
Given: a spring with k = 100 N/cm is
being stretched as shown
Find: the change in elastic energy when
the spring is stretched from (a) position 1
to position 2, (b) position 2 to position 3,
(c) position 1 to position 3
Solution:
Convert spring constant from N/cm to N/m
k  100 N/cm100 cm/m   10,000 N/m
© 2011 Cengage Learning Engineering. All Rights Reserved.
13-14
Example 13.3 – Elasticity Energy
Solution (continued):
1
1
change in elastic energy  EE  kx22  kx12
2
2
(a) EE 
(b) EE 
1
10,000 N/m 0.052  0  12.5 J
2
1
10,000 N/m 0.07 2  1 10,000 N/m 0.052  12 J
2
2
(c) EE 
1
10,000 N/m 0.07 2  0  24.5 J
2
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13-15
Conservation of Mechanical Energy
• The total mechanical energy of a system is
constant assuming
 negligible losses
 no work
 no heat transfer
KE  PE  EE  0
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13-16
Example 13.4 – Conservation of
Mechanical Energy
Given: a cart is rolling down
an inclined surface as
shown. The cart velocity at
point A = 2.5 m/s
Find: estimate H
Solution:
Neglecting rolling friction,
1
1
1
2
KE  mV22  mV12  m2.5 m/s   0
2
2
2
EE  0
PE  mgh   m 9.81 m/s 2 H




1
2
m2.5 m/s    m 9.81 m/s 2 H  0  H  0.318 m
2
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13-17
Thermal Energy
• Thermal energy (heat) transfer occurs
whenever there exists a temperature
difference within an object, between two
bodies, or between a body and its surrounding
• Heat always flows from a high-temperature
region to a low-temperature region
• Three modes of heat transfer
 Conduction
 Convection
 Radiation
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13-18
Thermal Energy
Unit:
In SI system, joule is the unit of energy
1 joule  1 N  m  1 kg  m2 /s 2
In U.S. Customary unit, thermal energy unit is given
in Btu and is related to mechanical energy through
1 Btu  778 lb  ft
1 Btu  1055 J
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13-19
Conservation of Energy – First Law of
Thermodynamics
• Energy is conserved
 it cannot be created or
destroyed
 energy can only change
forms
Q  W  E
where
Q   Qin   Qout
W   Wout   Win
E  net change in total energy of system
• For a system having a fixed
mass, the net heat transfer
to the system minus the
work done by the system is
equal to the change in total
energy of the system
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13-20
First Law of Thermodynamics
When it comes to energy,
The best you can do is to break even.
You cannot get more energy out of a system than the
amount you put into it
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13-21
Example 13.5 – Conservation of
Energy
Given: a heater puts 150 W (J/s) into the
water pot. Heat loss from the water pot to
the atmosphere is 60 W.
Find: the change in total energy of water
in the pot after 5 minutes
Solution:
Q  W  Qin  Qout   Wout  Win   E
W 0
150 J/s 300 s   60 J/s 300 s   E
E  27 kJ
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13-22
Power
• Power is defined as
 the time rate of doing work
 the required work (or energy) divided by
the time required to perform the task
work force distance 
power 

time
time
or
energy
power 
time
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13-23
Power
The faster you want the task done, the more power
you need to supply
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start
finish
More
power
13-24
Power
• Units
 Watts (W) in SI units
 Horsepower (hp) in U.S. Customary units
work force distance 
power 

time
time
Nm J

 W
s
s
lb f  ft
lb f  ft

and 1 hp  550
s
s
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13-25
Power
• General relationship between W and hp
 1 (lbf.ft/s) is slightly greater in magnitude than 1 W
lb f  ft
1
 1.3558 W  1.36 W
s

1 hp is slightly smaller than 1 kW
• 1 hp = 745.69 W  746 W
• Kilowatt hour (kWh) is a unit of energy – not power
 1 kWh = amount of energy consumed during 1 hr
by a device that uses 1 kW
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13-26
Example 13.6 – Power
Given: 30 people with an average mass of 61 kg (133
lbm) per person; vertical distance between 2 floors is 5 m
(16 ft)
Find: the power required to move these people between
2 floors in 2 seconds.
Solution:
power 
work

time
30 people  61

kg 
m
 9.81 2 5m 
person 
s 
 45,000 W
2s
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13-27
Example 13.6 – Power
Given: 1 million people use the stairs to go up one floor instead of
taking the elevator at work
Find: the amount of energy saved in a year (about 220 working days)
Solution:
From previous example, the min. energy required to move 30 people
between two floors is
work  power  time  45,000 W  2 s  90,000 J
The energy is equivalent to providing electricity to fifteen 100-W light bulbs
for 1 minute. For 1 million people,
 90,000 J  1 

1,000,000 persons 220 days 
energy savings  
 30 persons  day 
 66 109 J  660 GJ
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13-28
Example 13.7 – Power
Given: a person weighs 220 lb
Find: the power required to move
the person a vertical distance of 2.5
ft in 1 second.
Solution:
work 220 lb f 2.5 ft 
lb f  ft
power 

 550
 1 hp
time
1s
s
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13-29
Example 13.8 – Power
Given: an object weighs 800 N (179.85 lbf)
Find: the power required to move the object a vertical
distance of 4 m (13.12 ft) in 2 seconds. Express power
in SI and U.S. Customary units, and hp.
Solution:
In SI units
work 800 N 4 m 
power 

 1600 W
time
2s
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13-30
Example 13.8 – Power
Solution (continued):
In U.S. Customary units
work 179.85 lb f 13.12 ft 
lb f  ft
power 

 1180
time
2s
s
In horsepower

work
hp
 lb f  ft 
power 
 1180 

time
 s  550 lb f  ft
s

© 2011 Cengage Learning Engineering. All Rights Reserved.


  2.14 hp



13-31
Example 13.9 – Power
Given: a 2001 BMW 750iL weighs 4597 lbf, and the
engine in the car is rated at 326 hp at 5000 rpm
Find: is the claim that the car goes from 0 to 60 mph in
6.7 second justifiable?
Solution:
Assumptions: driver weighs 180 lbf, weight of car includes sufficient
gasoline for the test. We will convert speed and mass into appropriate units
ft
ft
 mi  1 h  5280 ft 
; V2  Vfinal   60 

  88
s
h  3600s  1 mi 
s

weight 4597  180 lb f
m

 148 slugs
ft
g
32.2 2
s
V1  Vinitial  0
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13-32
Example 13.9 – Power
Solution (continued):
the required work to go from 0 to 60 mph
2
work 1-2
1
1
1
 ft 
2
2
 mV2  mV1  148 slugs  88   0  573,056 lb f  ft
2
2
2
 s
the power requirement to perform this work in 6.7 seconds is
work  573,056 lb f  ft 
lb  ft

  85,530 f
time 
6.7 s
s




hp
 lb f  ft 
  155.5 hp
power  85,530 

 s  550 lb f  ft 
s 

Claim is OK!
power 
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13-33
Efficiency
• We use efficiency to express how well a
machine or a system is functioning
• All machines and engineering systems require
more input than what they put out
actual output
efficiency 
required input
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13-34
Power Plants
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12-35
Power Plant Efficiency
The overall efficiency of a steam power plant is
defined as
energy generated
power plant efficiency 
energy input from fuel
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13-36
Internal Combustion Engine Efficiency
power output
thermal efficiency 
heat power input as fuel is burned
Thermal efficiency of a typical gasoline engine
is approximately 25 to 30%
Thermal efficiency of a diesel engine is
approximately 35 to 40%
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13-37
Motor Efficiency
power input to the device being driven
efficiency 
electric power input to the motor
Efficiency of a motor is a function of load
and speed
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13-38
Pump Efficiency
power input to the fluid by the pump
efficiency 
power input to the pump by the motor
Efficiency of a pump, at a given operating speed,
is a function of flow rate and the pressure rise
(head) of the pump
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13-39
Pump Efficiency
• Function of a pump is to increase the pressure of a liquid
entering the pump and to transport liquid to higher
elevation if necessary
• Pumps are driven by motors and engines
• You find pumps in
 hydraulic systems
 fuel system of your car
 systems that deliver water
to a city piping network
 food processing
 petrochemical plants
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jacuzzi
pump
13-40
Cooling and Refrigeration System
• Main components of cooling and refrigeration
system
 vapor – compression cycle
•
•
•
•
Condenser
Evaporator
Compressor
Throttling device
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13-41
Cooling and Refrigeration System
pressure &
Temperature
increase
Heat transfer
Throttling device
refrigerant
Heat transfer
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13-42
Energy Sources, Generation,
Consumption
• The world’s growing demand for energy is among
one of the most difficult challenges that we face.
• As future engineers, you are faced with two
problems, energy sources and emissions;
solutions to these problems require innovative
approaches.
• Energy use per capita in the world has been
increasing steadily as the economies of the world
grow and the population continues to grow.
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13-43
Energy Sources, Generation,
Consumption
• Energy sources, generation, and consumption is
a global issue.
• The U.S. data given here is used to convey
important information to you.
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13-44
The U.S. Energy Consumption by
Source and Sector in 2008
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13-45
The U.S. Energy Production by Major
Source in 2008
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13-46
U.S. Electric Power Industry Net
Generation in 2007
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13-47
Coal
• 48.5% of all electricity generated in the U.S.
was created from coal.
• Coal-fired power plants burn coal in boilers or
steam generators to make steam. The steam
turns turbines that are connected to
generators to make steam.
• 93% of the coal mined in the U.S. is used for
generating electricity.
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13-48
Natural Gas
• The U.S. natural gas transportation network
consists of 1.5 million miles of mainline and
secondary pipelines.
• In 2008, the pipelines delivered more than 23
trillion cubic feet of natural gas to about 70
million customers.
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13-49
Heating Oil
• Heating oil is a petroleum product used to
heat homes in America, especially in the
Northeast.
• Heating oil and diesel fuel are similar in
composition with heating oil having more
sulfur than diesel fuel.
• U.S. IRS requires heating oil to be dyed red
because it is tax-exempt.
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13-50
Nuclear Energy
• There are two processes by which nuclear
energy is harnessed, nuclear fission and
nuclear fusion.
• Nuclear power plants use nuclear fission to
produce electricity.
• Nuclear fusion is the process by which the
sun’s energy is produced.
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13-51
Nuclear Fission Process
• The process splits the atoms of uranium and
releases more neutrons and energy in the
form of heat and radiation.
• The additional neutrons go on to bombard
other uranium atoms, and the process keeps
repeating itself, leading to a chain reaction.
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13-52
Nuclear Fission Process
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13-53
Nuclear Fusion Process
• In this process, energy is released when
atoms are combined or fused together to form
a larger atom.
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13-54
Percentage of Electricity Generated by
Nuclear Fuel during 1973-2008
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13-55
HydroPower
• Hydropower accounts for 6% of total U.S. electricity
generation.
• In 2008, it accounted for 67% of energy generation
of all the renewable energy sources.
• To generate electricity, water stored behind dams is
guided through water turbines that are connected to
generators located within the hydropower plant.
• Approximately 31% of total U.S. hydropower is
generated by the Grand Coulee Dam in
Washington.
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13-56
Solar Energy
• Solar energy starts with the sun at a distance
of 93 million miles from earth.
• The sun is a nuclear fusion reactor with its
surface temperature at approximately
10,000°F (5500°C).
• Solar energy that reaches the earth is in the
form of electromagnetic radiation consisting of
a wide spectrum of wavelengths and energy
intensities.
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13-57
Solar Energy
• The solar radiation could be divided into three
bands: the ultraviolet band, the visible band,
and the infrared band.
• The visible band comprises about 48% of useful
radiation for heating, and the near infrared
makes up the rest.
• Because the distance from the earth to the sun
changes during the year, the energy reaching
the outer atmosphere of the earth varies from
410 to 440 Btu/ft2
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13-58
Solar Energy
• Some of the solar energy passes through the
earth’s atmosphere is absorbed, some of it is
scattered, and some of it is reflected by clouds,
dust, pollutants, forest fires, volcanoes, or water
vapor in the atmosphere.
• Direct beam solar radiation is the radiation that
reaches the earth’s surface without being diffused.
• On a clear, dry day, 10% of direct beam radiation
can be reduced, and 100% can be reduced on a
thick, cloudy day.
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13-59
Solar Energy
The direct and diffuse radiation
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13-60
Solar Systems
• Solar systems convert solar radiations into
useful forms of energy such as heating water
or air.
• Two basic types of active solar heating
systems: liquid and air.
• Units for radiation data are
2
 kW/m /day
2
 Btu/ft /day (in U.S.)
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13-61
Schematic of a Solar Collector
Liquid is heated in a solar collector and
transported via a pump to a storage system
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13-62
Passive Solar System
• Passive solar systems do not make use of
any mechanical components such as
collectors, pumps, blowers, or fans to collect,
transport or distribute solar heat to various
parts of a building.
• Two types of passive solar systems
 Direct
 Indirect
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13-63
Photovoltaic Systems
• Photovoltaic System converts light energy directly
into electricity.
• The system consists of a photovoltaic array,
batteries, charge controller, and an inverter.
• The systems come in all sizes and shapes.
• The systems are classified into
 stand-alone systems
 hybrid systems
 grid-tied systems
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13-64
Photovoltaic Systems
• Stand-alone systems are not connected to a utility grid.
• Hybrid systems use a combination of photovoltaic
arrays and some other form of energy.
• Grid-tied systems are connected to a utility grid.
• Backbone of any photovoltaic system is the cell.
• Photovoltaic cells are
 combined to form a module and modules are
combined to form an array
 classified as crystalline, polycrystalline silicon, and
amorphous silicon.
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13-65
Examples of Photovoltaic Systems
(c) Space station; (d) a building rooftop
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13-66
Photovoltaic Plant
The Alamosa photovoltaic plant located in an area of 82 acres in
Colorado . It went on-line in 2007 and generates about 82 MW.
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13-67
Examples of Photovoltaic Materials
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13-68
Wind Energy
• Wind energy is a form of solar energy.
• Part of the kinetic energy created by air
movements can be converted into mechanical
energy and into electricity.
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13-69
Types of Wind Turbines
Schematics of two types of wind turbines:
a vertical axis and a horizontal axis
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13-70
Vertical Axis Turbines
• Vertical axis turbines can accept wind from any
angle and require light-weight towers and are
easy to service.
• The main disadvantage is its poor performance
because the rotors are near the ground where
the wind speeds are relatively low.
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Wind Turbines
• Most of the wind turbines in use in the U.S.
are horizontal axis turbines.
• Wind turbines are classified as
 small (< 100 kW)
 intermediate (< 250 kW)
 large (250 kW to 2 MW)
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Ethanol
• Ethanol is an alcohol that is made from the
sugars found in corn and barley, and other
sources such as rice, sugar cane, and potato
skins.
• Most of the ethanol used in the U.S. are
distilled from corn.
• The designation E10 is referred to a fuel that
is a mixture of 10% ethanol and 90%
gasoline.
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Biodiesel
• Biodiesel is a fuel that is commonly made
from vegetable oils or recycled restaurant
grease, and can be in diesel engines.
• The designation B20 is referred to a blend of
20% biodiesel and 80% petroleum diesel.
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Summary
• You should understand how different forms of
mechanical energy are defined.
• You should know how and when to use kinetic,
potential, and elastic energy in engineering analyses.
• You should know what is meant by internal energy
and heat.
• You should know how and when to use conservation
of energy to solve engineering problems.
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Summary
• You should clearly understand the definition of
power, its common units, and how it is related
to work and energy.
• You should know the basic definition of
efficiency and be familiar with its various
forms including thermal efficiency, COP,
SEER, and AFUE.
• You should have a basic understanding of
energy sources and generations.
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