PHYSICS AND TECHNOLOGY IV
INDTT 270
OLYMPIC COLLEGE
Name____________________
Date____________________
PHYSICS AND TECHNOLOGY IV
Jack Kinert
Bob Abel
Olympic College
Bremerton, WA
Copyright 2015
The Video version of the textbook is on-line. Go to
http://www.youtube.com/olympiccollege. Scroll down to where it says “Physics with Bob
Abel”. Click on the words (not the picture!) and you’ll find the entire list of Physics
videos in chronological order. Try to find the cameo appearance by Leonardo DiCaprio. I
can’t.
The textbook and the lab directions may be found on-line at
instructors.olympic.edu/psnsphysics/default.aspx. You’ll find the text under “PDF Files” and
the lab directions under “Links”.
Tutoring assistance is available Tuesdays and Thursdays after work in the Physics lab
(around 4:20). The main campus also has physics tutoring services.
Access Services coordinates accommodations for eligible students with disabilities and
works to ensure equal access to educational programs, services, and activities at Olympic
College. If you are in need of and/or eligible for such assistance, call them at 360-5757540 or visit their website at
www.olympic.edu/Students/StudentServices/AccessServices. Students may also take
advantage of the excellent Counseling Services on campus.
Acknowledgments
The authors gratefully acknowledge the assistance of Applied Physics Faculty and Staff
and many helpful Apprentices in the preparation of this text.
TABLE OF CONTENTS
I. INTRODUCTION
Grading
1
2
II: WAVES
A. LEARNING OBJECTIVES
B. INTRODUCTION
C. SIMPLE HARMONIC MOTION
PROBLEM SET 1
LAB 70: NATURAL FREQUENCY OF A
VIBRATING BODY
D. WAVES
PROBLEM SET 2
LAB 73: MEASURING PHASE DIFFERENCE
E. COMBINING WAVES - WAVE INTERFERENCE
PROBLEM SET 3
LAB 72: RESONANCE IN A HOLLOW SOUND TUBE
F. ELECTROMAGNETIC RADIATION
PROBLEM SET 4
LAB 75: THE ELECTROMAGNETIC SPECTRUM
G. LASERS
PROBLEM SET 5
LAB 78: LASER RADIATION
4
4
4
4
10
III: NUCLEAR RADIATION
A. LEARNING OBJECTIVES
B. INTRODUCTION
C. STRUCTURE OF THE ATOM
PROBLEM SET 6
D. NUCLEAR INTERACTIONS
PROBLEM SET 7
Lab 77 B: Calculating Radioactive Decay Products
E. HALF-LIFE OF RADIOACTIVE ISOTOPES
F. EFFECTS OF NUCLEAR RADIATION
PROBLEM SET 8
LAB 77C and D: NUCLEAR RADIATION
66
66
66
67
72
73
81
82
86
89
93
95
IV: ENERGY CONVERTERS
A. LEARNING OBJECTIVES FOR CHAPTER II
B. GENERAL DISCUSSION
HOW NUCLEAR PROPULSION
IN A SUBMARINE WORKS
PROBLEM SET 9
102
102
102
i
11
15
24
26
30
37
39
43
51
54
58
61
63
107
111
C. CONVERTING ELECTRICAL ENERGY
TO LIGHT ENERGY
PROBLEM SET 10
LAB 53: CONVERTING ELECTRICAL ENERGY TO
LIGHT ENERGY IN LAMPS
D. CONVERTING LIGHT TO ELECTRICITY
WITH SOLAR PANELS
PROBLEM SET 11
LAB 52: SOLAR PANELS
E. CONVERTING WIND ENERGY TO ELECTRICAL ENERGY
PROBLEM SET 12
LAB 51: CONVERTING WIND ENERGY TO
ELECTRICAL ENERGY
F. THERMAL ENERGY CONVERTERS
PROBLEM SET 13
G. THERMOELECTRIC GENERATOR
PROBLEM SET 14
LAB 55A & B: THERMOELECTRIC GENERATOR
PROBLEM SET 15
V: TRANSDUCERS
A. LEARNING OBJECTIVES FOR CHAPTER III
B. INTRODUCTION
C. FORCE TRANSDUCERS
PROBLEM SET 16
LAB 57: FORCE TRANSDUCERS (STRAIN GAUGES)
D. VIBRATION TRANSDUCERS (LAB 58)
PROBLEM SET 17
LAB 58: VIBRATION TRANSDUCERS
E. FLUID FLOW TRANSDUCERS
PROBLEM SET 18
F. FLUID PRESSURE TRANSDUCERS
PROBLEM SET 19
LAB 60: THE DIFFERENTIAL PRESSURE CELL
G. ELECTRICAL TRANSDUCERS
PROBLEM SET 20
ELECTROSTRICTIVE TRANSDUCERS
PROBLEM SET 21
H. LINEAR VARIABLE-DIFFERENTIAL TRANSFORMER
LAB 62: THE LINEAR VARIABLE
DIFFERENTIAL TRANSFORMER
I. THERMAL TRANSDUCERS
PROBLEM SET 22
ii
113
117
118
121
123
124
130
133
134
138
139
141
142
143
149
151
151
151
152
156
157
160
164
166
169
171
173
177
179
182
186
187
189
191
193
196
198
ANSWERS TO PROBLEM SETS
OSCILLOSCOPES
5-STEP METHOD OF SOLVING PHYSICS PROBLEMS
EQUATIONS
CONVERSION FACTORS & REFERENCE TABLES
SAMPLE UNITS FOR SOME COMMON VARIABLES
GLOSSARY OF TERMS
THE GREEK ALPHABET
iii
199
205
209
210
219
229
231
233
I. INTRODUCTION
TEXT: Physics and Technology IV, Kinert and Abel
OBJECTIVES: This course is designed to teach the basic principles of physics through hands-on experience
in the laboratory. Also, you will learn the proper use of technical equipment such as oscilloscopes, digital and
analog ammeters, wind tunnels, ohm meters, and fluid accumulators, transducers and solar cells. The following
four main subjects are covered in the fourth quarter:
“Waves” describes the manner in which energy is transported without the net movement of any mass. Sound,
vibrations, and electromagnetic vibrations, like light, are transported as waves.
“Nuclear Physics” is an introduction to the atom, the nucleus and nuclear radiation.
“Energy Converters” convert energy from one form to another..
“Transducers”, are devices used to measure operating parameters of technical systems to control and monitor
their performance.
LEARNING OUTCOMES AND ASSESSMENT METHODS: The learning outcomes and assessment
methods for this course are as described below:
LEARNING OUTCOMES
ASSESSMENT METHODS
1. Apply and explain basic and advanced techniques
for converting energy from one form to another.
1. Tests, experiments, lab reports, results reviewed
with student.
2. Apply and explain basic and advanced
measurement techniques for mechanical, fluid,
electromagnetic and thermal energy systems.
2. Tests, experiments, lab reports, results reviewed
with student.
3. Apply and explain the basic concept of waves and
radiation in mechanical and electromagnetic
systems.
3. Tests, experiments, lab reports, results reviewed
with student.
4. Solve advanced physics problems involving the
above concepts using the 5-step method for SI
and British unit systems.
4. Classroom discussion and problem solving, tests
and lab reports reviewed with students.
5. Using personal knowledge, physics concepts and
personal experience, critically analyze real world
physics problems.
5. Classroom discussion, tests and lab reports
reviewed with students.
6. Use written communications to demonstrate clarity
and content comprehension in writing.
6. Written lab reports, tests, reviewed with student.
7. Experiment and analyze data cooperatively to
develop peer instructing/learning skills.
7. Instructor group observation/critique during lab.
Graded supervisor’s report reviewed with student.
Self-assess and peer assess.
8. Attendance, classroom and laboratory observation
of individual effort, initiative, participation in lab
group and classroom discussions, completion of
assignments. Reviewed with student.
8. Behave responsibly and ethically as an individual
student and as a member of a team.
9. Develop self-assessment skills to modify
developing learning strategies.
9. Student keeps track of all grades on labs and test.
Review individual progress with instructor.
1
ASSIGNMENTS: This text provides the assignment and problems for each lab
experiment, and several pages for notes. Prior to each class, complete the assigned reading
in the text, and do the problems for that section. The instructor will provide an overview
and introduction to each lab in advance of each class.
The answers to all problems are provided in the back of the text. If you are unable to get
the correct answer, ask for clarification in the class or during the lab period.
At the completion of each experiment a supervisor’s report must be turned in for
evaluation. If possible, turn in this report at the end of the lab period. The report must
be turned in no later than the beginning of the next lab class.
TESTS: There will be four one-hour tests. This text may be used during tests. No other
texts or notes will be allowed. This includes material from the website of Dave Davis:
www.sinclair.net/~ddavis.
MAKE-UP WORK: You may arrange to take a test early if you know that you must be
absent on test day. If you miss a test, the make-up test must be taken within 7 calendar
days. Make-up tests will cover the same material as the classroom test, but the difficulty
will be greater. There will be no make-up labs.
LAB REPORT GRADES: The lab report forms in the text should not be removed
from the bound book; these will be your reference notes for tests. Work with your lab
partners to complete the answers to all questions and problems. The supervisor will
submit a report on a separate form, which will be evaluated in the following manner:
Experiment performed and report turned in .. 4 points
Accuracy and completeness of experiment .... 2 “
Questions & problems completed correctly.... 3 “
Neatness........... 1 “
10 points possible
All students in the supervisor’s group will receive the same grade, based on the
supervisor’s report. You will be the supervisor of 3 or 4 experiments. You will receive a
supervisor grade based on the criteria above and the overall group performance during the
lab and report completion phase (10 points possible). The three highest supervisor grades
will be used for your Supervisor grade (maximum 30 points).
QUARTER GRADE: The estimated point count (with 16 or 15 labs) is:
OR
16 labs x 10 pts:
160 points
15 labs x 10 pts:
150 points
Supervisor grade (3): 30 “
Supervisor grade (3): 30 “
4 tests x 50 pts each 200 “
4 tests x 50 pts each 200 “
Total
390 “
Total
380 “
2
Your grade will be determined by dividing your total points by the total possible. The %
value will be translated into the official grade according to the schedule below.
Decimal Point Grading Scale for INDTT Physics
Percentage
Point
Percentage
Point
Percentage
Point
Grade
Grade
Grade
Grade
Grade
Grade
95.5 – 100
4.0
83.8 – 85.1
2.8
71.8 – 72.4
1.6
94.5 – 95.4
3.9
82.5 – 83.7
2.7
71.1 – 71.7
1.5
93.5 – 94.4
3.8
81.8 – 82.4
2.6
70.3 – 71.0
1.4
92.5 – 93.4
3.7
81.0 – 81.7
2.5
69.5 – 70.2
1.3
91.8 – 92.4
3.6
80.3 – 80.9
2.4
68.5 – 69.4
1.2
91.0 – 91.7
3.5
79.5 – 80.2
2.3
67.5 – 68.4
1.1
90.3 – 90.9
3.4
78.1 – 79.4
2.2
66.5 – 67.4
1.0
89.5 – 90.2
3.3
76.6 – 78.0
2.1
65.5 – 66.4
0.9
88.5 – 89.4
3.2
75.0 – 76.5
2.0
64.5 – 65.4
0.8
87.5 – 88.4
3.1
74.2 – 74.9
1.9
63.5 – 64.4
0.7
86.5 – 87.4
3.0
73.4 – 74.1
1.8
59.5 – 63.2
0.6
85.2 – 86.4
2.9
72.5 – 73.3
1.7
3
CHAPTER II: WAVES
A. LEARNING OBJECTIVES FOR CHAPTER II
1. Describe wave motion in general and how waves transmit energy.
2. Describe the characteristics that are used to describe a wave, including wavelength, frequency,
period, amplitude, phase difference and speed.
3. Understand the difference between longitudinal and transverse waves.
4. Describe the characteristics of a harmonic wave and give examples of this type of wave.
5. Solve problems which involve the characteristics of a wave.
6. Identify workplace applications where waves and vibrations must be understood and
controlled.
7. Describe what is meant by interference of waves.
8. Understand what is meant by constructive and destructive wave interference.
9. Use the wave superposition principle to find the resultant of two or more waves interfering
with each other.
10. Determine the allowed frequencies and vibrations for standing waves.
11. Identify parts of the electromagnetic spectrum and their wavelength and frequency range.
12. Calculate the energy of a photon, given either its frequency or wavelength.
13. Understand the difference between laser light and other types of light.
14. Understand radiant power and power density.
15. Identify workplace applications of lasers.
B. INTRODUCTION
A wave is a traveling, periodic disturbance. Sound, light and earthquakes are all waves. Waves
travel through solids, liquids, gases and even the vacuum of space. In this chapter, some of their
properties are examined. We’ll start by discussing the source of all waves, periodic motion.
That will be followed by a description of wave characteristics, such as the period and frequency,
wavelength, amplitude, phase and velocity. We’ll find out how waves combine using the
principle of superposition, how standing waves form and how they lead to resonance. Finally
we’ll discuss electromagnetic radiation and you’ll learn the difference between blue light, red
light, x-rays and radio waves, the methods by which such radiation is produced, and how lasers
work.
C. SIMPLE HARMONIC MOTION
Simple harmonic motion (often abbreviated as SHM) is a very important type of vibrational
motion because so many natural phenomena are governed by it. SHM involves three
characteristics:
1) An object displaced from its equilibrium or rest position experiences a “restoring force”.
A force is applied to the object that tends to push it back toward its equilibrium position. For
example, if a pendulum bob is moved to the right and released, the force of gravity pulls it back
to the left. If the bob is displaced to the left, gravitational force pulls it back to the right.
However the bob is displaced, the force of gravity pulls is back toward its equilibrium position.
2) The restoring force is proportional to the displacement. Consider a spring with a Hooke’s
Law constant of one lb per inch (k = 1 lb/in). If the spring is stretched by one inch, it experiences
4
a restoring force of one lb. If the spring is stretched by two inches, the restoring force will be
two pounds. The restoring force is proportional to the amount of stretch.
3) The motion is periodic. It always takes the same amount of time to complete one period of
the motion. The Earth takes 365.25 days to revolve about the sun, and it does this every year.
The electrical current in your house completely reverses it direction and then changes back again
60 times per second. A resting heart completes a beat cycle, then starts again, about once every
second, so your heart beat is approximately periodic. As soon as we finish a day, another one
begins, and every one of them takes 24 hours. Electrons accelerating back and forth along a
conductive antenna one million times per second produce electromagnetic waves that bring us
live coverage of the Mariners' games on KIRO AM.
A one-time acceleration produces a pulse, but periodic motion, also known as a harmonic
oscillation, produces a periodic wave, a traveling, oscillating disturbance. Understanding waves
requires that we spend some time studying their production. The pendulum and the mass loaded
spring are two of the easiest sources of harmonic oscillation (periodic motion) that we can study
in the laboratory, so let's start with them.
The Simple Pendulum
A simple pendulum is shown in Fig. II-1. It consists of a mass (a “bob”) swinging from the end
of a string.


m
Figure II-1. A simple pendulum. The period of the swinging pendulum can be approximated
fairly accurately as long as a) all the mass can be considered to be at a distance  from the pivot
and b) the angle  is not too large.
As the bob swings back and forth it is constantly being accelerated by the gravitational force. As
it rises from the center position it decelerates, and as it falls toward the center position it
accelerates. The acceleration is constant: 9.8 m/s2 at sea level. The pendulum's energy is
constantly being transferred between kinetic energy and gravitational potential energy. At the
apex of its motion, where the potential energy is at its maximum, the speed of the mass drops to
zero, and its lowest point, where it attains it greatest speed, the kinetic energy is maximized. In
the absence of friction, this motion would continue forever, and every swing of the pendulum
would take the same amount of time to complete. Friction dampens the amplitude of the swing
until it finally stops, but until it stops, the period remains unchanged. That's right, the period of
the swinging pendulum does not depend on friction.
5
The simple pendulum, like all objects, tends to vibrate at a frequency depending on its own set of
characteristics. This frequency is called the natural frequency. The equation for the period (T)
of a simple pendulum depends only on its acceleration (g) and the length of the string ():


T  2
 2 
g
g
1/2
II-1
This is an approximation, but it works very well. It requires that the mass all be at a distance 
from the pivot, and that the maximum amplitude is not too far from the vertical position. That's
the definition of a simple pendulum. The smaller the amplitude of the swing the more accurately
eq. II-1 will predict its period. Its period is independent of the amplitude of its swing (as long as
it doesn't get too big) or the amount of mass hanging from the end of the string.
The frequency (f) is just the inverse of the period. The equation (from Physics 171) is
f
1
T
II-2
The units of frequency are typically hertz (Hz). One hertz is equal to one cycle per second
(remember that the period is the number of seconds per cycle).
Most pendulums are not simple. The sign in Fig. II-2 is swinging about its one remaining nail.
Its motion will be periodic and independent of its amplitude like the simple pendulum, but
calculating the period is much more complex because the mass is not all at one distance from the
pivot.
eN
Th
tS
igh
t
hif
Figure II-2. The mass distribution of a physical pendulum makes its period much harder to
calculate.
This type of pendulum is called a physical pendulum. We can still calculate its period using
calculus but, lucky for you, that is beyond the scope of this text.
Example II-1: A simple pendulum.
The pendulum in Fig. II-1 is 42 inches long with a mass of 40 g attached to the end. What is its
period of motion? What is the frequency?
6
 = 42 in
m = 40 g
T=?
f=?
We’ll need to use the English value for g, which is 32 ft/sec2. The units will work out to seconds
if we convert the length to feet:
 1 ft. 
  42 in.
  3.5 ft.
 12 in. 
The period is



3.5ft
T  2    2 
ft

g
 32 2
 s
and the frequency is the inverse of the period:
1/2
f






1/2
 2.08 sec
1
1
cycles

 0.481
 0.481 Hz
T 
sec
sec 
 2.08

cycle 

The units of period are generally just given as units of time, but we used the more descriptive
term “sec/cycle” to show that the inverse gives units of Hz. Note that neither the mass nor the
maximum displacement affects the period of the pendulum.
The mass loaded spring
A spring hangs from a mass in Fig. II-3. If we pull down on the mass and release it, the system
will undergo harmonic oscillation.
spring
constant
k
d
m
Figure II-3. A mass loaded spring system. The maximum displacement of the spring is given as
d.
Let's call the maximum displacement from equilibrium d. The period of oscillation only depends
on the amount of mass m and the spring constant, k. The spring constant is given by Hooke's
Law:
7
k
F
d
II-3
where F is the force acting on the spring and d is the distance it is being stretched or compressed.
The units of k are typically N/m = kg/s2. The equation for the period of oscillation is similar in
form to that of the simple pendulum. We'll make the same assumptions with the spring that we
did with the pendulum: we won't stretch or compress the spring too far and we'll assume that all
of the mass of the system is at one point. Obviously the lighter our spring is relative to the
oscillating mass, the better our approximation will be:
m
m
T  2
 2 
k
k
1/2
II-4
Once again we find that the maximum displacement does not enter into the equation.
Example II-2: The period and frequency of an oscillating spring mass system.
The spring in Fig. II-3 is displaced 7 cm from equilibrium when the 14 kg mass is hung from it.
What is the spring constant k?
d = 7 cm
m = 14 kg
k=?
14 kg 9.8 m2 
F mg
kg
N
s 

k 

 1,960 2 or 1,960
d
d
m
s
7 cm  1m 
 100 cm 
What would be the period and frequency of the oscillating mass spring system?
T=?


 14 kg 
m

T  2
 2
k
 1960 kg 


s2 

1/2
 0.531 sec
f=?
f
1

T
1
0.531
sec
cycle
 1.88
cycles
 1.88 Hz
sec
Note that once you’ve measured k with one mass, you can attach any other mass to the spring and
use the same value of k to calculate the period and the frequency. A larger mass, for example,
would stretch the spring further and still give you the same spring constant value.
Forced Vibrations
The vibrations of a ship’s propellers are transmitted to the turbine. These vibrations can cause
misalignment and result in damage to the turbine and shaft. The pounding of a heavy air
8
compressor in a building can be transmitted by waves to delicate electronics in a nearby
laboratory. In these examples, the propeller vibrations and air compressor vibrations are forced
vibrations.
In general, when an outside force is applied at regular intervals to an object, the object will be
forced to vibrate at the frequency of the applied force. Forced vibrations can be good or bad. For
example, many musical instruments use vibrating strings or reeds to produce a sound. String
instruments, like a piano or guitar, use forced vibrations to amplify the sound. The sounding
board of the piano and the body of the guitar are forced to vibrate at the frequency of the string.
Reed instruments, like clarinets, make good use of forced vibrations. The vibrations from the
reed force columns of air to vibrate along the body of the clarinet, amplifying the sound.
Electromagnetic oscillations that arrive at a stereo speaker can not be heard, but they force the
speaker cone to vibrate. The vibrating cone sets large amounts of air into motion. The air in
motion makes up the longitudinal sound waves - a series of compressions and rarefactions - that
travel to your ear.
Forced vibrations in mechanical systems usually are unwelcome. That is because forced
vibrations coming from other sources cause the machine to vibrate at unwanted frequencies,
resulting in overheating, excessive wear, misalignment of moving parts, reduced performance
and breakdown.
Rubber mounts or shock absorbers isolate mechanical systems from forced vibrations by
absorbing and damping the source vibrations.
Forced vibrations are also important in thermal systems. You will remember that temperature is a
measure of the average kinetic energy of the molecules of the material. A technician can increase
the temperature of a material by forcing the molecules to vibrate faster. One example is
induction heating, where an induction coil is used to cause random electrical currents to flow in a
material. The currents, flowing through the resistance of the material, induce heat directly into it.
This way the heating can be quick, localized and easily controlled. This process is used in the
metal industry to solder, braze, weld, anneal and harden various metal alloys.
The Tacoma Narrow Bridge disaster is a classic example of what happens when a forced
vibration matches the natural frequency of an object. In this famous example, the blowing wind
caused the forced vibrations. These vibrations matched the natural frequency of the suspended
bridge. As the forced vibrations gave more and more energy to the bridge, the bridge swayed
back and forth with larger and larger amplitude until it tore itself apart.
Figure II-4. The original Tacoma Narrows Bridge, “Galloping Gertie” responds to forced
vibrations due to the strong winds in the area.
9
PROBLEM SET 1: SIMPLE HARMONIC MOTION
1. Find the length of a simple pendulum which will have a period T = 1 sec.
2. A 1 kg mass is attached to the end of a spring with a spring constant of 20 N/m. What are the
a) period and b) frequency of vibration?
3. Find the a) period and b) frequency of a vibrating structure that completes 3000 cycles in 6
sec.
4. A simple pendulum has a length of 0.40 m. What are its a) period and b) frequency?
10
OVERVIEW LAB 70: NATURAL FREQUENCY OF A VIBRATING BODY
When energy is added to vibrating systems like the pendulum and spring, they tend to oscillate at
their natural frequency.
Vibration means that the object is displaced from its rest position and reaches some maximum
displacement. Then the cycle is reversed with the object moving through the rest position to the
other side.
Vibration at the natural frequency continues indefinitely, unless energy leaves the system. A
spring may use energy to do work, or frictional forces remove energy from the system. When
energy leaves the system, you get a damped vibration. In a damped vibration, the object's
displacement from rest decreases with each vibration cycle. Eventually, the vibration stops.
What sets the natural frequency of oscillation? The answer depends on the type of vibrating
system with which we are dealing. We will consider two systems, the simple pendulum and the
mass-spring system.
THE PENDULUM
An object suspended so that it swings back and forth freely is called a pendulum. You are
probably familiar with pendulums from seeing them in grandfather clocks. A simple pendulum
consists of a mass which hangs from a string or other flexible material. In a simple pendulum we
always assume that the string has no mass (usually a good approximation, since the string has
very little mass compared to the pendulum bob).
When a pendulum is displaced horizontally from its rest position, it begins to oscillate at its
natural frequency. This frequency is given by the equation below.


T  2
 2 
g
g
1/2
where g = acceleration due to gravity
 = the length of the pendulum string
T = period of oscillation
What are pendulums used for? In old clocks, pendulums are used to measure time. In more
recent applications, pendulums are used to detect motion. In such a system, the pendulum is
suspended in a magnetic field. When the pendulum begins to move, it generates a voltage signal.
An example of such a system is the seismograph, which is used to detect the motion of the earth's
surface (very useful in an earthquake).
11
THE SPRING
A vertically-suspended spring with a weight attached will vibrate at its natural frequency when
displaced from its equilibrium position. When the spring vibrates, its natural frequency is given
by the equation below:
m
T  2  
k
1/ 2
where m = mass on end of spring
k = spring constant
T = period of oscillation
The spring constant, k, is a number that describes the stiffness of the spring. k has units of force
divided by length such as lb/in or N/m. The stiffer a spring is, the higher the spring constant will
be.
Springs are used in many mechanical systems. They are often used to control the periodic
motion of mechanical devices. Shock absorbers in cars are a form of spring. Vibration mounts,
which are used in machine shops to stabilize precision machines, are a form of spring also.
Springs chosen for these applications have a natural frequency different from where they are
used. Because of this, they stabilize the machine on which they are mounted.
12
LAB 70: NATURAL FREQUENCY OF A VIBRATING BODY DATE___________
OBJECTIVES:
SKETCH:
TABLE 1 PENDULUM VALUES
MASS
OF
PENDULUM
BOB
PENDULUM WITH 20-cm CORD
TIME
FOR 10
SWINGS
(sec)
MEAS’D
PERIOD
Tm (sec)
CALC’D
PERIOD
Tc (sec)
PENDULUM WITH 40-cm CORD
TIME
FOR 10
SWINGS
(sec)
MEAS’D
PERIOD
Tm (sec)
CALC’D
PERIOD
Tc (sec)
100 g
200 g
CALCULATIONS: Calculate Tc for each length and record in Table 1.
TABLE 2
Tc  2

g
SPRING VALUES
HANGING
MASS
FORCE
DUE TO
MASS
F (N)
DISTANCE
STRETCHED
 (m)
0.5 kg
F1 =
1 =
1.0 kg
F2 =
2 =
TIME FOR
20 CYCLES
(sec)
CALCULATIONS: Find TC and record in Table 2. Period:
Spring Constant: k = F/ = (F2 - F1)/(2 - 1)
13
MEAS’D
PERIOD
Tm (sec)
m
Tc  2 
k
CALC’D
PERIOD
Tc (sec)
1/ 2
LAB 70: ANALYSIS
1. Did changing the mass of the pendulum affect the period?______
2. Compare Tm with Tc for the 100 g mass. Calculate the % difference for each length:
3. What might be the sources of error in the pendulum experiment?
4. How could you slow down a grandfather clock that was running fast?
5. What change would there be in the period of a pendulum used in this lab if it were oscillating
on the moon? (The acceleration due to gravity on the moon is about one-sixth of that on
earth.)
6. Calculate the frequency for each pendulum length. Remember that frequency, f, is the
reciprocal of the period, T.
7. How did changing the hanging mass on the spring affect the frequency of oscillation?
8. Find the % difference between the measured and calculated periods of the spring trials:
9. The mass of the spring was neglected in the calculated Tm. The effective mass is actually
greater than the value used in the calculation. If the actual effective mass had been used,
would the calculated value of Tc be larger or smaller?
10. How could you increase the period of oscillation for a spring? (Give two methods.)
11. Calculate the frequency of each spring trial
14
D. WAVES
Wave motion begins with the vibration of some source. A wave is a propagating disturbance
through some medium in which there is no net motion of the medium. A sound wave, for
example, is a propagating pressure disturbance, in which molecules are smashed into each other,
but return to their original positions. We observe water waves on the ocean moving toward the
shore, but there is no net displacement of the water. If we send a wave pulse along a taut rope,
we observe that each section of the rope is in the same position after the pulse leaves as before it
arrived. Waves are only traveling energy; there is no net motion of matter.
Mechanical Waves and Electromagnetic Waves
Mechanical waves are a propagating disturbance through some kind of matter, such as air,
water, rock or steel. Sound is a mechanical wave, which is why, in space, no one can hear you
scream. Electromagnetic waves do not require matter to propagate, in fact matter slows
them down. Electromagnetic waves propagate fastest through a vacuum, which is why the sun’s
rays can reach us from 93 million miles away through the desolate blackness of space. The light
that we see, plus radio waves, microwaves, infrared rays, x-rays, ultraviolet rays and gamma rays,
are all electromagnetic, or “e-m”, waves. Let’s state this one more time: a mechanical wave
requires a medium to transport it. Electromagnetic waves do not.
Harmonic Waves
A single disturbance produces a pulse. A continuous wave is produced by a series of
disturbances. A harmonic wave, a special type of continuous wave, is produced by simple
harmonic motion. Its shape is that of a very special, smooth wave called a “sinusoidal” (sine)
wave that repeats itself over and over again. Two continuous waveforms are shown in Fig. II-5.
The top one is a harmonic wave and the other is a non-uniform wave.
period
amplitude
Harmonic waves have uniform period and amplitude…
… other waves don’t.
Figure II-5. Harmonic and non-uniform wave shapes.
The wave shape shown at the top of Fig. II-5 has the form of a sine wave. Notice how smooth
and regular it appears. The high peaks and low peaks (or troughs) are always the same distance
above or below the center line (neutral position). Notice that the distance between adjacent high
peaks (or adjacent low peaks) is always the same, all along the wave. And notice how the shape
repeats itself exactly.
15
By contrast, the adjacent peaks of the non-uniform wave at the bottom are different distances
above and below neutral position, and the distance between the adjacent peaks varies randomly.
If a cork bobbing up and down on the pond were to produce this type of wave, you can imagine
that its motion would be rather irregular.
Because sinusoidal waves are made up of smooth, repeatable patterns, we can use them to define
important wave characteristics. These characteristics are discussed below.
Wave Characteristics
Waves, particularly harmonic waves, are described in terms of their amplitude, period,
frequency, wavelength, phase and speed. We describe each in turn below.
Amplitude
Before the vibrations that cause a mechanical wave get started, the molecules in the medium are
relatively still. They are in what is called a neutral position. During the wave motion, the
maximum distance that molecules move away from this neutral position is called the amplitude
(A) of the wave. The amplitude of a harmonic wave is the distance of a crest above the neutral
position as well as the distance of a trough below the neutral position. It is initially controlled by
the vibrating source: the larger the vibrations at the source, the larger the initial wave amplitude.
The amplitude of a wave is an important characteristic because it is related to how much energy
the wave is transporting.
The wave amplitude is listed as “A” in Fig.’s II-6 and II-7. This is different than the peak-topeak Voltage we measure on an oscilloscope; the wave amplitude is only half of that value.
Period and Frequency
The number of seconds that it takes each wavelength to pass by a fixed point along the wave is
called the period (T). The units are those of time (seconds, minutes, hours, days, etc.). The
period is the amount of time it takes for one cycle to pass a given point, so the units can also be
given as seconds per cycle, etc. The number of cycles that pass by any fixed position along the
path of wave travel, each second, is called the frequency (f). Frequency is the inverse of the
period, as shown in Eq.. II-2. Another definition for frequency is that it is the number of cycles
(n) per time (t):
f
n
t
II-5
The frequency of the wave is the same as the frequency of the vibrating source. The units most
commonly used to measure frequency are cycles (complete waves) per second, also known as
hertz (Hz). The period and amplitude of a harmonic wave are shown in Fig. II-6. Here the
period marks the distance between two crests of the wave, but it can be measured between any
two points that cover one complete cycle of the wave.
16
T
1
Displacement from Equilibrium
T = 8 sec - 2 sec = 6 seconds
A
f = 1/T = (1 cycle)/(6 sec)
= 0.167 cycles/sec
= 0.167 hertz
= 0.167 Hz
0
0
2
4
6
time (seconds)
8
10
A
-1
Figure II-6. Example of a harmonic (sinusoidal) wave showing the period T and the amplitude
A.
Examples II-3, 4 and 5 show how to calculate the wave frequency and wave period.
Example II-3: Wave and Frequency and Period
The rower in the boat below notices that 3 wave crests pass by every 2 seconds.
a) What is the frequency of the wave?
n = 3 cycles
t = 2 sec
f=?
f
n 3 cycles
cycles

 1.5
 1.5 hertz  1.5 Hz
T
2 sec
sec
b) What is the period of the wave?
T
1

f
1
sec
 0.667
 0.667 sec
cycles
cycle
1.5
sec
17
Note that we could have listed the frequency units as Hz instead of cycles/sec, but this way it’s
easier to make sure the units work out correctly.
Example II-4: Frequency and Period of Oscilloscope Sine-Wave
A sine wave trace on an oscilloscope looks like that of Fig. II-6. The “time/div” setting is 0.05
sec/div and the distance of one complete cycle (as shown) is 6 divisions.
a) What is the period of the wave?
# divisions = 6 div
time/div = 0.05 sec/div
The period is determined by multiplying the number of divisions by the time per division (we
did this in Physics 171):
sec 
 time 

T  # divisions
  6 div  0.05
  0.3 sec
div 
 division 

b) What’s the frequency of the wave?
1
1
cycles
f 
 3.33
 3.33 Hz
sec
T
sec
0.3
cycle
Example II-5: Period of a Radio Wave
A local FM radio station broadcasts at a frequency of 98.5 megahertz (MHz).
a) What is the period of the wave in seconds?
f = 98.5 MHz = 98.5 x 106 Hz = 98.5 x 106 cycles/sec
T=?
T
1

f
1
98.5 x 10 6
cycles
sec
 1.02 x 10 -8
sec
 1.02 x 10 -8 sec
cycle
Note that we didn’t convert the frequency to scientific notation (9.85 x 107 Hz), we just re-wrote
“Mega” as 106. It reduces the opportunities for conversion errors. Also note how small the
period of an electromagnetic wave such as an FM signal is – about 10 billionths of a second!
Wavelength
The length of one cycle of a wave is called its wavelength. It is denoted by the Greek symbol
“”, pronounced “lambda”. Its units are those of distance (ft, m, in, cm, etc.). Since it is the
length of one cycle, we sometimes include that information as well (ft/cycle, m/cycle, etc.). We
measure the period by plotting the wave vs. time; we measure the wavelength  by plotting the
wave displacement vs. distance, as shown in Fig. II-7.
18

1
Displacement from Equilibrium
 = 8 m - 2 m = 6 meters
A
0
0
2
4
6
distance (meters)
8
10
A
-1
Figure II-7. The wavelength is the length of a wave’s cycle and the amplitude is the maximum
displacement from the undisturbed position.
Example II-6 shows how to calculate wave amplitude and wavelength for water waves passing by
a boat.
Example II-6: Wave amplitude and wavelength
Water waves pass by the 16-foot fishing boat shown in Example II-3. There are 4 complete
wavelengths spanning the length of the boat. The rower notices that the vertical distance between
adjacent crests and trough is about one-fourth as high as the boat, which is 28 inches high.
a) What is the amplitude of the wave? Let the height of the boat be hboat and the height of
the wave be hwave:
hboat = 28 in.
hwave = hboat/4
A=?
The amplitude of the wave is half its height:
A
h wave 1 h boat 1  28 in. 

 
  3.5 in.
2
2 4
2 4 
b) What is the wavelength? Let the number of cycles be n and the length of the boat be boat:
boat = 16 ft
n = 4 cycles
=?

 boat
16 ft
ft

4
 4 ft
n
4 cycles
cycle
19
Phase
The phase of a wave describes the location of a point along one cycle of a wave. The location of
a point along a single wave cycle can be described in terms of fractions of a cycle or, since it
repeats itself like a circle (or since it’s a sine wave), we can use angular terms like radians or
degrees. One complete cycle is equal to 2 radians or 360 degrees. The sine wave in Fig. II-8
below is plotted as a function of degrees. The location of the wave is known as the phase angle.
One quarter of a wave is 90º; one half of a wave is 180 º. Two complete cycles are (2)(360 º) =
720 º.
Figure II-8. The phase of a wave can be given in terms of fractions of a cycle, radians or, as
shown here, in degrees.
Phase is particularly useful when comparing two waves whose peaks don’t match up. The phase
difference between two waves has the same units as phase. Alternating current and voltage each
are represented by sine waves. When a simple circuit contains only resistance, the sine waves that
represent the current and voltage are "in phase". This means that the crest (peak) on one wave
matches the crest (peak) on the other. Also, the trough on one wave matches the trough on the
other.
Adding other electronic components to the circuit - such as capacitors or inductors - generally
causes the current and voltage waves to be "out of phase". The difference between similar points
on the waves is called the "phase difference". Phase difference can be measured in fractions of a
wavelength or cycle. Since one wave cycle equals 3600, phase difference usually is measured in
degrees.
Fig. II-9 shows the current and voltage oscillations for a simple AC circuit. The current reaches
its amplitude (maximum value) an eighth of a cycle (45º) before the voltage. The current is said
to "lead" the voltage (or the voltage “lags” the current) by 45 º. We say the phase difference is
45º.
20
2
Displacement from Equilibrium
1.5
1
Current (I)
0.5
Voltage (DV)
0
0
90
180
270
360
450
540
630
720
-0.5
-1
-1.5
-2
Phase Angle (degrees)
Figure II-9. Voltage and Current in an AC circuit are often out of phase. In this case the phase
difference is one-eighth of a cycle, or 450. The current is said to lead the voltage or, equivalently,
the voltage lags the current, by 450.
It’s not hard to measure phase difference for two waves with the same wavelength (or period,
depending on whether you’re looking at the wave in terms of distance or time). Find the length
of one cycle and see what fraction of that cycle two peaks or troughs differ by. The two waves in
Fig. II-9 are 8 divisions long, and their peaks differ by one division. You don’t have to use the
peaks if you don’t want to, just the same position on a cycle (the phase) for the two waves.
In the case of an AC circuit, the phase difference, or phase angle, must be determined to
determine the total power. Whereas the power in a DC current is given by P = V·I, the power in
an AC circuit is given by
P  V  I  cos
II-6
where  is the phase angle between the voltage and the current. We’ll practice measuring phase
difference in Lab 73.
Wave Speed
The wave speed or velocity (v) of a traveling harmonic wave (sine wave), like a sound wave, or a
water wave, or an electromagnetic wave, is the product of the wave frequency and the
wavelength:
v  f
II-7
Since f=1/T, we can also write the velocity in terms of the period and the wavelength:
v

T
II-8
The units work out to be distance over time, which is what we need for velocity.
21
Example II-7: Wavelength of a sound wave
The “A” note above middle “C” has a frequency of 440 Hz. If the speed of sound is 330 m/s,
what is the wavelength of “A”?
f = 440 Hz = 440 cycles/sec
v = 330 m/s
=?
m
330
v
s  0.75 m  0.75 m
v  f

 
cycles
f
cycle
440
s
or about two and a half feet.
Example II-8: Wavelength of a radio wave
What is the wavelength of the 98.5 MHz FM radio station in Example II-5? The speed of a radio
wave (or any electromagnetic wave, including light) is 3 x 108 m/sec.
f = 98.5 MHz = 98.5 x 106 Hz
v = 3 x 108 m/s
=?
m
v
s
 
 3.05 m
cycles
f
98.5 x 10 6
s
3 x 10 8
Effect of the Medium on Wave Speed
The speed of a pulse or wave in a medium depends on two characteristics of the medium:
1. The velocity increases with the strength of the coupling between adjacent particles. In a
stretched string, the greater the tension, the faster the wave will travel. In a spring, the higher the
Hooke's law constant, the faster a wave will travel. A sound wave will move faster in solids than
liquids, and faster in liquids than gases. Sound travels faster in an elastic solid, such as steel, than
in an inelastic one, such as lead. If the temperature of a gas is increased, the molecules interact
more strongly with each other, and the speed of sound will increase.
There is an increase in the speed of sound as air temperature increases. At 200C, the speed of
sound in air at sea level is 343 m/s. The speed increases about 0.59 m/s per each C0 increase.
2. The velocity decreases with the greater inertia of the particles. A wave will travel more slowly
in a heavy string than in a light string under the same tension. A wave will travel more slowly in
a dense spring than in a light spring, if the Hooke's law constants are the same. If materials have
the same strength of coupling between molecules, sound will travel more slowly in the more
dense substance. If two gases have the same temperature, sound will travel more slowly in the
more dense gas. For example:
The density of air at 00C is 1.29 grams/m3, and the density of helium at 00C is 0.178 g/m3. The
velocity of sound in air at 00C is 331 m/s, and the velocity in helium at the same temperature is
965 m/s.
22
Electromagnetic waves that travel through matter (for example, light travels through glass or
water) are slowed down from their vacuum speed of 3.0 x 108 m/s (186,000 miles/sec). The
amount by which it is slowed depends on the medium. This phenomenon produces the effect
known as refraction.
Longitudinal Waves and Transverse Waves – the relative direction of the disturbance
The disturbance that characterizes sound waves causes particles to be displaced back and forth in
the direction that the wave is traveling, also known as the direction of wave propagation. These
are longitudinal waves: the wave displacement is parallel to the direction of propagation.
Waves produced by plucking a guitar string produce a displacement along the string
perpendicular to, or transverse to the direction in which the wave propagates. Transverse waves
are waves whose displacement (or vibration) is perpendicular to the direction of
propagation.
A transverse wave is shown in Fig. II-10a and a longitudinal wave is shown in Fig. II-10b. Fig.
II-10b shows the density variations of a traveling sound wave, alternating between dense
(crowded) regions of molecules and regions where the density is less than that of the air before
the wave came through. Note that the two waves shown in a and b are of the same wavelength
and are in phase: regions of high density in the longitudinal wave correspond to peaks in the
transverse wave, and regions of low density in the longitudinal wave correspond to troughs in the
transverse wave.
amplitude
a) transverse wave
Period
b) longitudinal wave
Figure II-10. A transverse wave (a) and a longitudinal wave (b, denoted by dark and light regions
of high and low density). The waves have the same wavelength and are in phase.
23
PROBLEM SET 2: WAVES
1. A smooth, repeating wave that has a sinusoidal shape is called a _____________________
wave.
2. The particles of a rope vibrate in a direction perpendicular to the direction of a wave traveling
along the rope. This kind of wave is called a _______________ wave.
3. Air molecules next to a loudspeaker vibrate in a direction along the path of the sound energy.
This kind of wave is called a
________________ wave.
4. Mechanical waves ____________ (require, do not require) a medium containing matter for
propagation, while electromagnetic waves ________________ (require, do not require) a
medium.
Select a word or letter from the list below which best fits the description given for problems 510.
a. frequency
b. period
c. wavelength
d. amplitude
e. speed
5.___ distance between two adjacent crests on a transverse wave.
6.___ distance between two adjacent high pressure regions in a sound wave.
7.___ the distance from a peak of a transverse wave to the neutral position.
8.___ the time it takes for a cycle to pass a fixed position.
9.___ the rate at which complete cycles pass by a fixed location.
10___ the product of wavelength and frequency.
11. Waves are sometimes drawn with a horizontal axis that is marked in degrees. One complete
wave or cycle equals _______degrees.
12. Determine the phase difference - in degrees - for each pair of waves below:
a)
24
displacement from equilibrium
2
1
Phase Angle
(degrees)
0
0
90
180
270
360
450
540
630
720
-1
-2
b)
displacement from equilibrium
2
1
Phase Angle
(degrees)
0
0
90
180
270
360
450
540
630
720
-1
-2
13. A radio station broadcast an FM signal at a frequency of 92.5 Megahertz. The FM signal
travels through the air at 3 x 108 m/s. Determine a) the frequency, b) the period and c) the
wavelength.
25
OVERVIEW LAB 73: MEASURING PHASE DIFFERENCE
In this lab, you will measure phase shift. You have learned that a wave passes through 360
degrees when it completes one cycle.
In electronic systems, technicians are able to troubleshoot circuits by knowing what the phase
shift should be between various points in the circuit. Phase shift gives information about the
difference between input and output signals in control circuits. But one of the most important
uses of phase shift is in power measurement.
In DC electrical circuits the power is equal to the product of the circuit's voltage and current.
P=VxI
However, when working with AC circuits, there is another factor that must be taken into account.
It is quite possible that the voltage and current in an AC circuit are out of phase (there is a phase
angle between them). In this case, the product of voltage and current must be multiplied by a
correction factor, which is always less than one. In other words, the true power is less than the V
x I value when the current and voltage are not in phase. Mathematically, the correction factor is
the "cosine" of the phase angle.
Calculator Exercise:
1. Be sure your calculator is set to the "degree" mode (not radians or grads). Enter the value
0 (zero) in your calculator. This will represent I and V in phase (zero degrees out of phase).
Now press the key for cosine, which is marked "cos". The display shows a value of 1. This
tells us that when the current and voltage are in phase, the power is the full value of V x I.
2. Now enter 45 in the calculator. This represents I and V 45o out of phase (1/8 of a cycle). Press
the cos key. The display shows the correction factor of 0.707. This means that the power will
only be seven-tenths of the full V x I value.
3. Now enter 90 in the calculator. This represents I and V 90o (one-fourth of a cycle) out of
phase. When I is greatest, V is zero. When V is greatest, I is zero. Part of the time the voltage
is positive while the current is negative. Press the cos key. The display shows a factor of zero,
indicating that there was a complete cancellation of the power produced.
Mathematically, this power equation is expressed:
P = (V x I) cos
where P = power
V = voltage
26
I = current
 (“theta”) = phase angle between the voltage and current
cos is commonly known as the power factor. The case where V and I are out of phase by 45
degrees is shown in Figure II-9. Knowing the phase angle in power circuits is critical when it is
necessary to get the maximum power out. In this lab you will find the phase angle in a sine wave
using a resistor-capacitor (RC) circuit, a function generator, and an oscilloscope.
To find the phase difference between the input and output signals: determine the number of grid
spacings between where the input signal crosses the horizontal centerline and where the output
crosses the centerline. Calculate  for each trial:

# of grid spacings of phase difference
 360 
# of grid spacings per cycle

27
LAB 73: MEASURING PHASE SHIFT
DATE________
OBJECTIVES:
SKETCH OF EQUIPMENT:
PHASE SHIFT: # OF GRID SPACES: 1 _________ 2 _________ 3 _________
# OF DEGREES:
1_________
2_________ 3_________
4
-4
4
-4
Oscilloscope for phase shift #1
4
4
-4
-4
4
4
-4
-4
Oscilloscope for phase shift #2
Oscilloscope for phase shift #3
28
LAB 73: ANALYSIS
1. Draw two sine waves where the phase between the two waves is:
a. 90 degrees
b. 180 degrees
4
4
-4
-4
-4
4
4
-4
2. Assume that you have an AC circuit in which V = equals one volt and I equals one amp.
Knowing that P = V·Icos in AC circuits, find out how much power is provided when:
a. = 0 degrees, P = _________
c.  = 90 degrees, P = ________
b.  = 30 degrees, P = _________
d.  = 360 degrees, P = _______
3. Explain what you think the answer to 2-d means.
4. You have a power circuit that gives maximum power (0 degrees phase between voltage and
current). A technician plays with the circuit and now you measure the phase shift as 60
degrees. What is the percentage of power that you can now get from the circuit?
5. Consider the circuit used in this experiment. What could you change in the circuit that would
change the phase angle between the input and output phase angle? Give at least two factors.
29
E. COMBINING WAVES - WAVE INTERFERENCE
When two of more of the same type of waves pass through the same point at the same time, the
medium responds to each wave. The overlapping of waves at a point is called interference.
When waves interfere, their displacements at any one point combine arithmetically, that is, they
add. The overall effect is either greater or less than the effect of any one of the waves alone,
depending on the relative displacements (remember, the displacements can be positive or
negative). If the overall displacement is greater than each of the individual waves, the
interference is called constructive interference. If the overall displacement is less, the result is
described as destructive interference. Surfers ride the high crest caused by constructive
interference of water waves. When water waves interfere destructively, there are no crests to ride,
and they’re forced to luau. The process of ADDING the two waves together algebraically when
they coincide is called superposition.
Whenever two or more waves meet and overlap, the resultant effect equals the sum of the effects
caused by each wave alone. The sketch in Fig. II-11 shows two identical waves in an elastic
medium. The amplitude of the resultant pulse is the sum of the two waves. In this case, since the
two waves are identical and in phase, the summed wave has twice the amplitude (2) of that of the
individual waves (1).
2
displacement from equilibrium
Sum of the Two Waves
(Constructive Interference)
1
Original
Waves (2)
Phase Angle
(degrees)
0
0
90
180
270
360
450
540
630
720
-1
-2
Figure II=11. Two identical waves in phase, with an amplitude of 1, add at each point along the
cycle to form a summed wave with an amplitude of 2. Since the combined wave has a greater
amplitude than each of the individual waves, this is constructive interference.
Figure II-12 shows two waves that are 180º out of phase. At each point along the x-axis, the two
individual waves add up to zero. The summed wave is coincident with the x-axis. This is an
example of complete destructive interference.
30
displacement from equilibrium
2
1
Phase Angle
(degrees)
0
0
90
180
270
360
450
540
630
720
-1
-2
Figure II-12. Destructive interference of two wave pulses.
Superposition is used to find the resultant wave produced by the combination of individual
waves. The method for combining the waves is simple: at each point along the x-axis, add the
displacements of the individual waves. An example of this is shown in Fig. II-13. Waves #1 and
2 are the individual waves; their sum is given by the heavy solid line. The two individual waves
are 45º out of phase.
2
Sum of Waves #1 and 2
Displacement from Equilibrium
1.5
1
Wave #1
0.5
Wave #2
0
0
90
180
270
360
450
540
630
720
-0.5
-1
-1.5
-2
Phase Angle (degrees)
Figure II-13. Using superposition to add two overlapping waves.
At x=0º, wave #1 has a displacement of 0.75 and wave #2 has a displacement of 0. Their net
displacement (sum) is 0.75. At 45º, wave #1 has a displacement of 1 and wave #2 has a
displacement of 0.75. Their sum is 1.75, as shown on the graph. The waves are added at each
point the x-axis, using the vertical grids as a guideline. What is the displacement of the summed
wave at 270º? The displacement of wave #1 is about -0.75 while that of wave #2 is -1, so the net
displacement is -1.75. Check other points along the graph to make sure you know how
superposition works. We’ll only use two waves to show superposition, but be aware that any
31
number of waves can be present in one area, and all their displacements at each point must be
added to determine the resultant wave.
Knowing about waves and how they interfere can be important in the workplace. Interference is
especially useful in light wave technology (optics) and sound wave technology (acoustics). The
transmission of vibrations through machinery can produce significant damage if multiple waves
interfere constructively. You’ve seen waves in a pool or bathtub reflect off a wall and interfere
with incoming waves. Sometimes the constructive interference between two waves heading in
different directions can produce very large amplitude waves or pulses. The same effect can occur
when sound waves (vibrations) are traveling through machinery. Often one mechanical system
must be isolated from others on a ship or submarine. Acoustical “live” spots and “dead” spots in
an auditorium result from constructive and destructive interference, respectively, between
multiple acoustic waves. The walls of recording booths are designed to reduce the amount of
reflected sound to minimize wave interference.
When light waves of similar wavelength (color) interfere they can produce a series of bright and
dark regions called an interference pattern. The bright regions (sometimes called bands) are the
result of constructive interference, while the dark bands (where troughs meet crests) are the result
of destructive interference.
There are many times when such interference of light plays an important role in the workplace.
Machinists use optical flats to determine the flatness of machined parts. An optical flat is a piece
of very flat glass about 3/4 of an inch thick. It is placed on the machined surface, and a light is
directed toward the surface. As shown in Fig. II-14, some of the light is reflected by the
machined surface; some is reflected by the glass.
incident light
rays
reflected light
rays
machined
surface
optical flat
Figure II-14. An optical flat. Some of the incident light rays are reflected off the surface of the
flat, while some are reflected off the machined surface. The reflected waves interfere with each
other and the resulting wave depends on the difference in their travel paths. In this sketch, the
incident light rays are in phase, while the reflected light rays are completely out of phase,
resulting in dark bands (destructive interference).
The difference in path length between the waves reflecting off the optical flat and those reflecting
off the machined surface will interfere. The reflected waves in Fig. II-14 interfere destructively,
and will produce very little, if any reflected light. Elsewhere, where the machined surface is
further from or closer to the optical flat, the difference in path length will produce different
amounts of interference. Some of it will be constructive and produce bright regions of reflected
light. Machinists observe the light and dark regions produced and determine where the machined
32
surface is not flat. Fig. II-15 is an example of an optical flat, with light and dark regions of
interference.
Figure II-15. Interference patterns indicate the flatness of machined parts.
Musicians use interference to tune their instruments. Two sources of sound waves that are nearly
at the same frequency will interfere constructively and destructively in a regular pattern. That
might happen when two strings of a guitar tuned to almost the same note are played. The listener
then hears a sound that varies regularly in intensity. These variations are called beats. For
example, if sound wave #1 is 99% as long as sound wave #2, their peaks will match every 100
cycles. In between, the summed will slowly decrease in amplitude and then increase. This is the
beat. The time between peaks will get longer and longer as the difference in the wavelengths
becomes less and less; the time between beats increases as the two wavelengths, or notes,
become more in tune.
Musicians use one musical instrument or an audio generator as a standard. When the same note is
sounded on a second instrument, beats will occur if the frequencies are close but not the same.
The musician adjusts the second instrument until the beats slow down and finally disappear. The
instrument is now in tune. The number of beats per second tells the musician how far the two
instruments are out of tune.
Standing Waves
A standing wave is formed when two like harmonic waves of equal amplitude and frequency,
traveling in opposite directions, interfere with one another. They may be transverse waves, such
as waves in a string, or longitudinal waves, such as sound waves. The result is a wave that
appears to stand still in space, that is, certain regions exist wherein displacement is quite large,
while other regions exist wherein there is little or no displacement.
Standing waves can be generated by holding a slinky at both ends and vibrating one end so that
the largest displacement occurs at the center. A snapshot of the slinky, where the displacement
has reached its maximum positive value, would look like the graph in Fig. II-16.
33
1
1st Harmonic:
 = 2
0
0
1
-1
Figure II-16. The first harmonic standing wave for a vibrating system with both ends fixed.
Since both ends are fixed, they are forced to be nodes, the points where the displacement is zero.
The incident and reflected waves are in phase in a standing wave. The anti-node, the point of
maximum displacement, is in the center. This configuration is known as the first harmonic; the
number of the harmonic in this case corresponds to the number of anti-nodes in the system.
Vibrating one end of the slinky faster will eventually get you to the second harmonic. As we
can see in Fig. II-17, the second harmonic has two anti-nodes.
1
2nd Harmonic:
=
0
0
1
-1
Figure II-17. The second harmonic of a vibrating system with fixed ends.
The first harmonic is a half wave, while the second harmonic is a full wave. Vibrating one end
of the slinky still faster produces higher frequency, shorter waves. As the waves get smaller, you
can eventually fit one and a half, and then two complete waves between the stationary ends.
These are the 3rd and 4th harmonics, respectively, and they are shown in Fig. II-18.
1
1
4th Harmonic:
 = /2
rd
3 Harmonic:
 = 2/3
0
0
0
0
1
-1
1
-1
Figure II-18. The 3rd and 4th harmonics of a vibrating system with fixed ends.
Allowed frequencies and wavelengths of standing waves
34
We can determine the frequency (f) and wavelength () of any harmonic in a standing wave
system given the speed of the traveling wave (v) and the length of the system ().
For the system with both ends fixed (a fixed end system), that is, the ends have to be nodes, the
first harmonic occurs when there is only one anti-node, the second occurs when there are two, the
third occurs when there are three, etc. (see Fig.’s II-16, 17 and 18). The number of wavelengths
fitting along the length  are ½, 1, 1½ and 2, respectively, so we see that we increase the number
of wavelengths by ½ for each harmonic. The relation between the length of the system, the
wavelength and the harmonic number (n) for a system with fixed ends is
n
II-9
2
We can always solve for the wavelength if we are given the length of the system and the
harmonic number. If we know the speed of the wave, we can determine the harmonic frequency:

f
v

which is just Eq. II-7.
Example II-9: Standing waves on a vibrating string.
A 4.5 meter long string vibrates at its 3rd harmonic, as shown in Fig. II-18. The speed of the
traveling wave is 36 m/s.
a) What is the wavelength for the 3rd harmonic?
 = 4.5 m
v = 36 m/s
n=3
=?
From Eq. II-9,

n

2

2 24.5 m 

 3m
n
3
b) What is the frequency of the wave?
v
f 

m
s  12 cycles  12 Hz
m
s
3
cycle
36
Some vibrating systems, such as an open sound tube, have anti-nodes at either end. They are
called open end systems. We can determine the harmonics in the same way as with the fixed
end system, but in this case the ends are anti-nodes and the first harmonic has one node, the
second has two nodes, the third has three nodes, etc. Eq. II-9 still works for this system. Each
higher harmonic holds an additional half wavelength. The first harmonic for an open end system
is shown in Fig. II-19.
35
1
1st Harmonic:
 = 2
0
0
1
-1
Figure II-19. The first harmonic for an open-end system. The ends have to be anti-nodes, and
the harmonic number is equal to the number of nodes.
36
PROBLEM SET 3: WAVE INTERFERENCE
1. Draw the resultant wave produced by the two waves shown below:
displacement from equilibrium
2
1
Phase Angle
(degrees)
0
0
90
180
270
360
450
540
630
720
-1
-2
2. Two identical waves are completely in phase (0º phase shift) in the configuration below,
so that they look like one wave. Draw the resultant wave produced by these identical, in
phase waves.
displacement from equilibrium
2
1
Phase Angle
(degrees)
0
0
90
180
270
360
450
-1
-2
37
540
630
720
3. The wave speed along a fixed-end slinky system is 10 m/s when the slinky is stretched to a
length of 4 meters. What are the wavelength and frequency of the first three harmonics?
4. An open-ended sound tube creates standing waves when both ends are anti-nodes. What are
the wavelength and frequency of the 2nd harmonic of a standing wave in a 30 cm long tube if the
speed of sound is 330 m/s?
38
LAB 72 OVERVIEW: RESONANCE OF SOUND WAVES IN HOLLOW TUBES
Resonance in a vibrating system occurs when the frequency of some external vibration matches
the natural frequency and phase of the system on which it is operating. This results in an increase
in the amplitude of the combined waves due to wave superposition. Resonance can occur when
sound is transmitted down a hollow, open tube of uniform cross section and the reflected wave is
in phase with the transmitted wave. This is known as an open end system, where the anti-nodes
are at either end of the tube. The shape of the first resonant wavelength is shown in Fig. II-19.
In this lab, you will use a function generator and speaker to create resonance in an open tube.
You will use a microphone as a collection device, and an oscilloscope as a measuring device to
find the first four resonant frequencies of a sound signal. At these frequencies you will observe
increased transmission of sound through the tube. You will then find the theoretical values for
resonant frequencies and compare them to the measured values. The sound waves will be
generated at a known frequency using a function generator and a speaker. The sound wave will
enter the tube at one end, propagate down the tube, and striks the microphone at the far end.
Using the function generator you can change the frequency of the sound being sent through the
tube. By doing this you will see that there are some frequencies at which the tube efficiently
transmits sound – the resonance frequency. For the majority of frequencies, however, the tube
transmits almost no sound.
You learned that sound is transmitted in waves. Each repetition of a wave pattern is called a
cycle. The distance between consecutive peaks in a set of waves is called the wavelength (λ).
The number of waves passing a point per unit of time is called the frequency. The function
generator/speaker combination emits uniform sound waves at a particular frequency.
For an open tube of length , resonance will occur for the following wavelengths:
First resonance: λ1 = 2 or  = ½λ1
Second resonance: λ2 =  or  = λ2
Third resonance: λ3 = (2/3) or  = (3/2)λ3
Fourth resonance: λ4 = /2 or  = 2λ4
Study the values just given. Notice that resonance occurs whenever the length of the tube is
equal to a multiple of half-wavelengths of a sound wave traveling through it. The frequency is
related to the wavelength by the following equation:
f = vs
λ
Where: f = frequency of the wave in Hz
vs = velocity of the wave in m/s
λ = wavelength of the sound wave in the tube in meters.
39
EXAMPLE: Find the first four resonant frequencies of an open tube 2.0 meters long.
Given:
vs = 331.4 m/s (speed of sound in air at 0o C)
 = 2.0 m (open tube length)
λ1 = 2,
f1 = vs  2 = 331.4m/s  (2)(2)m  83 Hz
λ2 = ,
f2 = vs   = 331.4m/s  2m  166 Hz
λ3 = (2/3), f3 = vs  (2/3) = (331.4)m/s (2/3)(2)m  249 Hz
λ4 = /2,
f4 = vs  /2 = (331.4)m/s  (2 m)/2  331 Hz
Calculations for LAB 72:
1. Find the measured period for each trial, according to the TIME/DIV setting of the
oscilloscope. Record in Table 1.
2. Find the measured frequency for each trial. Remember that the frequency is the reciprocal of
the period. Record in Table 1.
3. Using the scaled thermometer, measure the temperature in your classroom. Record the
temperature in Table 2.
4. The speed of sound in air changes with temperature. Sound speed increases 0.59 m/s for every
increase of one C0. Find the speed of sound in air for your classroom using the value of 331.4
m/s for the speed of sound at 00 C. (Add 0.59 m/s times the C0 above zero.
5. Find the wavelength of each resonant frequency for both tubes, using the equation: λ=vs
f
λ1 = vs
λ2 = vs
λ3 = v3 λ4 = v4
f1
f2
f3
f4
Record your results in Table 2 for measured resonant wavelengths (λmeas).
6. Find the theoretical resonant wavelengths, using the equations below:
λ1 = 2
λ2 = 
λ3 = (2/3)
λ4 = ½ 

Record your answers in Table 2 for calculated resonant wavelengths (λcalc).
40
LAB 72 RESONANCE OF SOUND WAVES
OBJECTIVES
TABLE 1
SKETCH
TUBE LENGTH : SHORT_____________ LONG______________
RESONANCE #
SHORT TUBE
Date____________
Function
Generator
fgen(Hz)
# Horizontal
Div. For One
Wave Cycle
Time/Div
Control
Setting
Measured
Period
T (sec)
Measured
Frequency
f (Hz)
1
2
3
4
LONG TUBE
1
2
3
4
TABLE 2
Resonance
#
o
Temp =
C
Calculated speed of sound: vs =
SHORT TUBE
LONG TUBE
Measured Resonant
Wavelength
Calculated Resonant
Wavelength
Measured Resonant
Wavelength
Calculated Resonant
Wavelength
λmeas (m)
λcalc (m)
λmeas (m)
λcalc (m)
1
2
3
4
41
LAB #72 ANALYSIS
1. Predict the wavelengths of the first four resonant frequencies for a tube 50 cm long. Do the
same for a tube 4 meters long. Show your work.
2. Given that a tube has its first resonant frequency at 1000 Hz, find the length of the tube. Do
the same for 1000 Hz as the third resonant frequency. Show your work. TAIR = 250 C.
3. Explain how resonance in a sound tube is similar to resonance in other systems. Use what
you know about waves and offer a few examples of other resonant systems.
4. Explain what would happen to your results if you were to run this experiment at the
temperature of 0o C. Explain what would happen if you were to run this experiment at 100o
C. Would the resonant frequencies increase or decrease? What about the resonant
wavelengths?
5. What other evidence, besides the oscilloscope trace, did you have to show that you had
reached a resonant frequency?
42
F. ELECTROMAGNETIC RADIATION
Radiation is movement of energy away from an energy source. There are three basic types of
radiation: 1) mechanical waves such as sound waves or water waves, 2) electromagnetic (e-m)
waves, in which energy is transmitted by fluctuating electromagnetic fields, and 3) particle
emission, which is produced by nuclear reactions. We’ve looked into mechanical waves and
we’ll discuss particle emission in the next chapter on nuclear radiation, but in this section we will
describe electromagnetic radiation.
Electromagnetic radiation (e-m) is wave energy carried by fluctuating electric and magnetic
fields. It is generated by accelerating charges, such as electrons oscillating back and forth along a
transmission antenna, by the spontaneous energy decay of bound electrons (to be explained in
this chapter), and by the emission of energy from atomic nuclei (to be discussed in the next
chapter).
Electromagnetic radiation includes AM and FM radio waves, TV broadcast waves, radar,
microwaves, infrared waves (heat radiation), visible light, ultraviolet waves (“UV’s”), x-rays and
gamma rays. The only difference between these types of waves is their wavelength. The only
difference between AM radio waves and visible light, for example, is that the AM radio waves
are much longer (or their frequency is much lower). All electromagnetic radiation travels at
the same speed in a vacuum: 3 x 108 m/s (186,000 miles/sec). This is known as the speed of
light; it is the universal speed limit. Nothing that we currently know of can go faster. It is
so special that we give this speed a special name: c. We write the velocity equation for e-m
radiation using c instead of v, as in Eq. II-7:
c = f
II-10
Our eyes are attuned to a very tiny fraction of the e-m spectrum, which is the range of
wavelengths (or frequencies) over which e-m waves occur. Part of this is due to the size of our
eyes; it wouldn’t be practical to have eyes big enough to receive wavelengths the size of AM
radio waves. Another factor is the processing ability of the brain: the bigger the range of
wavelengths (frequencies) the eyes can receive, the more complex they and the brain have to be.
Evolution appears to have maximized our visual efficiency. The wavelengths we are most
sensitive to are the wavelengths that are most likely to be received by our eyes. The sun’s
emission spectrum peaks at a wavelength of about 5.50 x 10-7 m, and this is the frequency that
our eyes are most sensitive to.
A chart of the electromagnetic spectrum is shown in Fig. II-20 as a function of wavelength and
frequency. Note that frequency (on the left) decreases as wavelength (on the right) increases, as
one should expect from Eq. II-10. Typical units are given next to the frequency and wavelength
powers of 10. For example, a wavelength of 10-10 meters is typically referred to as 1 angstrom =
1 Å. The type of light we call each section of the spectrum (like gamma rays or x-rays) are listed
in the middle. Notice what a small section of the total spectrum we can see, and this scale is
listed in powers of ten! On a linear scale the section of the spectrum we call visible light would
be even smaller. It is no wonder we use so many instruments to gather in the information
available from the rest of the spectrum.
Figure II-20 sketches out the electromagnetic spectrum on the following page. Can you find the
section covering the visible spectrum?
43
Our eyes distinguish between the different frequencies (wavelengths) of the visible spectrum
through color discrimination. That’s how our brains tell us we’re seeing different frequencies.
Different people have different talents for discriminating between frequencies (colors). Those
who have the best talents can become fashion designers or match the old paint on their houses.
Those who can’t have to take up other jobs, like teaching physics, and should never be allowed to
try to match the old paint on their house.
The longest wavelengths (lowest frequencies) in the visible range are those of red light. We
can’t see longer wavelengths – “infrared” means “below red”. In order of increasing frequency
(or decreasing wavelength), the colors are roughly red, orange, yellow, green, blue, indigo and
violet. The first names of the main colors of the spectrum form an acronym: “Roy G. Biv”. If
44
you combine all the colors of the spectrum, you get white light. Sun light is actually composed
of all the colors of the spectrum.
Some sections of the spectrum overlap, like ultraviolet and x-rays. This is because the sections
are dependent upon where the e-m radiation originates. The radiation in the overlap regions is
exactly the same; only the source varies. For example, x-ray and gamma rays in the overlap
come from different sources; x-rays come from the electrons in orbit about the nucleus while the
gamma rays of the same f and λ come from inside the nucleus.
Characteristics of e-m radiation
Electromagnetic radiation shares many of the same characteristics of any other harmonic wave,
such as amplitude and the relation between wave speed, wavelength, frequency and period. The
following are two examples to remind you of these relations.
Example II-10: The wavelength of an e-m wave.
A 1000 kHz radio wave (KOMO 1000 AM) is transmitted to a receiver at 3x108 m/s. Find the
wavelength.
f = 1000 kHz = 1000 x 103 cycles/sec
c = 3 x 108 m/s
=?
c
 
f
m
s
 300 m
cycles
3
1000 x 10
s
3 x 10 8
Light slows down as it passes through matter. When this occurs, its wavelength is reduced, but
its frequency remains unchanged. Visible light slows down very little in air – almost not at all
– as we see in the next example.
Example II-11: An e-m wave traveling through glass.
Red light (f = 5 x 1014 Hz) has a wavelength of 600 x 10-9 m = 600 nanometers (nm) in air and
400 nm in glass. What’s the wave speed in a) air and b) glass?
f = 5 x 1014 cycles/sec
air = 600 x 10-9 m
glass = 400 x 10-9 m
a) vair = ?


cycles 

-9
8 m
v air  f air   5 x 1014
 600 x 10 m  3.00 x 10
s 
s

which is the same value we gave as c, the value in vacuum. The speed of light in air is
actually just a tiny fraction slower than in a vacuum.
b) vglass = ?
45


cycles 

-9
8 m
v glass  f glass   5 x 1014
 400 x 10 m  2.00 x 10
s 
s

We see that the speed of light in glass is 2/3 the speed of light in air! This gives rise to the
phenomenon known as refraction – the light appears to “bend” as it enters the glass.
Wave-Particle characteristics of electromagnetic radiation
Electromagnetic radiation has the properties of both waves and particles. This strange mix of
characteristics can be described mathematically by quantum mechanics, but visualizing it is
beyond the ability of even the greatest physicists. Electromagnetic radiation can be described by
its wavelength and frequency, but it also has particle characteristics, such as momentum and a
kind of individuality not observed in other types of waves. The particle associated with
electromagnetic radiation is known as a photon. We can speak of red light as a wave or as a
series of photons with a given wavelength and energy. It’s pretty weird stuff, even for the people
who figured it all out.
Energy of electromagnetic energy
The energy of an e-m wave, or the energy of a photon (E), is proportional to its frequency (f):
E  hf 
hc

II-11
where h is a constant, known as Planck’s constant. The value of Planck’s constant is
approximately 6.626 x 10-34 J·sec, a pretty small number. We can use Eq. II-10 to write the
photon energy in terms of the wavelength as well. Long e-m waves are less energetic than short
e-m waves (Fig. II-21).

long wavelength, low frequency, low energy

short wavelength, high frequency, high energy
Figure II-21. The energy of an electromagnetic wave is proportional to its frequency, and
inversely proportional to its wavelength.
Example II-12: Energy of a gamma ray
A gamma ray with a wavelength of 1 x 10-12 m is moving in air. Find the energy of this photon.
Assume the speed in air is the same as the speed in vacuum.
46
 = 1 x 10-12 m
E=?
hc
E


6.626 x 10

m

J  s  3 x 10 8 
s

 1.99 x 10 -13 J
-12
1 x 10 m
- 34
The energy of a gamma ray photon is very small but it can penetrate steel or lead.
Example II-13: Energy and frequency of microwaves
A potato is cooked on a paper napkin in a microwave oven. The microwaves have a wavelength
of 0.03 m. Find the a) frequency and b) energy of the microwaves.
 = 0.03 m
a) f = ?
c
f 

m
s  1 x 1010 cycles  1 x 1010 Hz
m
s
0.03
cycle
3 x 10 8
b) E = ?


cycles 

-24
E  hf  6.626 x 10-34 J  s  1 x 1010
  6.63 x 10 J
s 

You can see from these two examples that microwaves have much, much smaller energy than
gamma rays. The reason that microwaves can quickly cook food is that photons of this
wavelength are more readily absorbed by the water molecules in the food, which gives them
more thermal energy. Microwaves heat up the water in an object, which is why your plate stays
relatively cool while the food cooks.
You can increase the e-m energy by increasing the frequency of each photon, or you can add
more photons. The total energy of electromagnetic energy (Etotal) of a set of photons of a given
energy is the product of the total number of photons (n) and the energy of each photon (Ephoton).
Etotal  nEphoton
II-12
Example II-14: The power of the sun
The sun puts out about 3.26 x 1026 W of power – that’s 3.26 x 1026 J of energy each second. If
we approximate each photon as having a wavelength of 5.50 x 10-7 m, about how many photons
are emitted by the sun each second?
47
Etotal = 3.26 x 1026 J
t = 1 sec
l = 5.50 x 10-7 m
n=?
The energy of each photon is
hc
Ephoton 


6.626 x 10

m

J  s  3 x 108 
J
s

 3.61 x 10-19 J  3.61 x 10-19
7
photon
5.50x10 m
- 34
The number of photons is
E total  nEphoton 
n
E total

Ephoton
3.26 x 10 26 J
 9.03 x 10 44 photons
J
-19
3.61 x 10
photon
which is about a billion trillion trillion trillion photons each second.
Units for e-m wavelengths
Common units used for describing the wavelength of e-m radiation are
a)  in meters
b)  in angstroms where 1Å = 10-10 meter
c)  in nanometers where 1 nm = 10-9 meter
d)  in microns where 1 m = 10-6 meter
Angstroms are commonly used for ultraviolet light, x-rays and visible light, nanometers are more
common for longer wavelength e-m radiation, and microns are normally used for describing
infrared wavelengths.
48
E-M radiation sources: blackbody radiation vs. electron absorption and emission
Incandescent light bulbs use a different mechanism to produce light than fluorescent light bulbs.
The incandescent bulbs produce light by heating a tungsten filament to high temperatures. The
electrons and atoms in the filament are vibrating over a range of high speeds, and these rapidly
accelerating charges consequently emit electromagnetic waves with a wide range of frequencies.
The white light we see from the bulb is actually the superposition of the entire visible spectrum
of photon frequencies, plus frequencies above and (mostly) below the visible range. The jiggling
atoms and electrons in cooler objects also emit blackbody radiation, but over a lower range of
frequencies. If you use a rheostat to dim the bulb, it not only burns less brightly, but the light is
redder, so fewer photons are being produced and over a lower frequency range. This is known as
blackbody radiation and it produces a continuous spectrum of e-m frequencies.
The fluorescent bulbs operate via electron emission and absorption of photons. Electrons bound
to atoms can be thought of orbiting the atomic nucleus. Their orbits differ from the orbit a planet
can follow around a star, however. Quantum mechanics tells us that electrons are confined to
certain orbits at certain discrete distances from the nucleus. These distances depend on the atom
they are orbiting. An electron in a higher orbit has more energy than one in a lower orbit.
Electrons do not change orbits by moving from one to another – they simply stop being in one
orbit and start being in another, so they are never caught in between orbits. This seems pretty
weird, but that’s quantum mechanics for you. Electrons move to higher orbits by absorbing
electromagnetic energy (photons) – this is called electron absorption. Electrons move to lower
orbits by emitting photons – this is called electron emission. A cartoon of the process is
provided in Fig. II-22.
photon is
absorbed
photon is
emitted
Figure II-22. The electron in the atom on the left absorbs a photon and moves to a higher energy
level. The photon energy has to be equal to the energy difference between the two orbits. On the
right side, the electron moves back down to the lower energy level. The photon emitted must be
of the same energy, and therefore of the same frequency and wavelength, as the absorbed photon
on the left. The energy of the photon emitted by an electron moving down to a lower orbit is
equal to the difference in energy level between the two orbits.
Since the electron orbits are discrete, the energy they absorb or emit to move between two orbits
(also called energy levels) is always the same for those two particular orbits. We can determine
the initial and final orbits of an electron moving between two orbits by the frequency, or color, of
the photon that is absorbed or emitted.
49
The fluorescent light bulb emits light by exciting electrons in a phosphor atom and letting them
drop to a lower energy level by emitting a discrete light frequency. Manufacturers add different
types of atoms to the phosphor coating to produce enough different frequencies to resemble a
continuous spectrum, but if you spread out the light using a spectroscope, you can see that the
fluorescent light is actually putting out only a few discrete wavelengths. That’s why the
fluorescent light doesn’t look quite the same as the incandescent bulb light.
In this lab we’ll excite different atoms electrically (through collisions with electrons) and observe
the frequencies (colors) they emit when the bound electrons move back down to lower energy
levels. These discrete colors act as fingerprints that allow us to determine the identity of the
atoms that produce them, because each type of atom has its own peculiar set of electron orbits
(energy levels).
50
PROBLEM SET 4: ELECTROMAGNETIC RADIATION
1. Match the following:
a. _____ energy unit
b. _____ λ
c. _____ high frequency EM
d. _____ invisible light
e. _____ h = 6.626 x 10-34 J-s
1.
2.
3.
4.
5.
wavelength
ultraviolet
joule
Planck's constant
x-rays
2. Which color photon has the largest energy?
3. Which color photon has the longest wavelength?
4. As the photon energy increases, what happens to λ and f?
5. Find the wavelength in Å of
a) λ = 0.7 μm
b) λ = 0.4 x 10-4 cm
c) λ = 555 nm
6. Given that infrared e-m radiation has λ between 10-4 and 10-6 m, find the frequencies that
correspond to these wavelengths.
51
7. The FM band of radio broadcasting waves uses electromagnetic radiation with frequencies
between 85 and 110 Megacycles/sec and wavelengths of about 3 meters.
a) Find the frequency for a signal of wavelength 3.25 meters.
b) Find the energy associated with this frequency.
8. When light passes from air into another medium such as glass, its wavelength and speed
change, but its frequency remains the same. For example, when light of wavelength 550 nm
and frequency 5.45 x 1014 Hz passes from air into thin glass, the wavelength changes and the
speed is reduced to 2 x 108 m/s, while the frequency remains the same. Find the wavelength
in the thin glass, if the speed is 2 x 108 m/s and the frequency is constant.
52
9. The energy of a particular photon of light is found to be 5 x 10-19 J and it is moving in air.
Find the wavelength and frequency of this light.
10. Ultraviolet light of wavelength 6.75 x 10-8 m is traveling in air. Find the energy of a photon
of this light.
53
OVERVIEW
LAB 75: THE ELECTROMAGNETIC SPECTRUM
Electromagnetic radiation is wave energy carried by electromagnetic fields. You are familiar
with many types of such radiation. Visible light, infrared, ultraviolet, x-rays, gamma-rays, and
microwave radiation are all electromagnetic in nature.
The energy of an electromagnetic wave is given by
E
where:
hc
 hf

h = Plank's constant = 6.626 x 10-34 J·sec
λ = wavelength (in m)
f = frequency of e-m radiation in hertz
c = speed of e-m radiation in vacuum = 3 x 108 m /sec
The speed of e-m radiation in vacuum (c) is one of the fundamental constants in nature. It is
commonly called the speed of light. It is the same for all forms of e-m radiation, regardless of
the wavelength.
Another fundamental equation for e-m radiation describes the relationship between the frequency
and the length of each e-m wave. That relationship is written:
c = f
where:
c = speed of e-m wave in a transparent medium (m/sec)
λ = wavelength of e-m wave (m)
f = frequency of e-m wave (Hz)
In a transparent medium, the speed of light is slightly different for each frequency; the dispersion
of colors in the rainbow is caused by this difference. However, for the effects studied in this
experiment, the speed of light in air may be taken to be constant at 3 x 108 m/sec for the entire
visible spectrum. Note that the relationship between f and λ is such that as f increases, λ
decreases and vice versa. It is said that frequency and wavelength are inversely related.
In this lab, you will use a spectroscope to view the types of energy emitted by various gases. The
type of energy given off by any type of radiator is called the radiator's spectrum. The spectrum
for each type of chemical element is unique.
In nature, two types of spectra exist. One is the type you observe when looking at electrified
gases through the spectroscope. This type is called a line spectrum because you see distinct lines
of different colors when viewing the gases through the spectroscope. This type of spectrum is
produced when electrons orbiting the nucleus of an atom are de-excited to a lower orbit.
The other type of spectrum is called a continuous spectrum. This is because, when looking at
one through a spectroscope, you see a continuous band of colors. In actuality, the bands are
really many single lines in the spectrum. However, since the lines are so close together, the
spectrum appears continuous. This type of spectrum is usually emitted by radiators that are
54
heated and made to glow. You will view such a spectrum when looking at an incandescent light
bulb.
Scientists use different types of spectroscopes to study the chemical nature of stars, interstellar
gases and planetary atmospheres. They compare their spectra to the spectrographs of known
elements. In this manner they are able to identify their chemical composition. They do that in
much the same way you will identify chemical gases in a tube.
In this lab you will observe both line and continuous spectra. You will compare what you see
through the spectroscope with a chart of spectra from known elements. Then you will identify
the line spectrum for various gases. The centerline wavelength of the various colors in the
visible spectrum are given in the Table above.
Color of light
wavelength (Å)
wavelength (m)
Red
6700
6.7 x 10-7
Orange
6200
6.2 x 10-7
Yellow
5700
5.7 x 10-7
Green
5200
5.2 x 10-7
Blue
4700
4.7 x 10-7
Indigo
4400
4.4 x 10-7
Violet
4100
4.1 x 10-7
55
LAB 75: THE ELECTROMAGNETIC SPECTRUM
OBJECTIVES:
DATE________
SKETCH:
DATA TABLE 1: SPECTRA OF GAS SAMPLES
Wavelength in Angstroms (Å)
3500 4000 4500 5000 5500 6000 6500
750
Hydrogen
Helium
Nitrogen
Mercury
Neon
DATA TABLE 2: SPECTRUM OF INCANDESCENT LIGHT
Wavelength in Angstroms (Å)
3500 4000 4500 5000 5500 6000 6500
56
LAB 75: ANALYSIS
1. In the chart below, calculate the frequencies for the colors in Fig. 7 ( in the directions)
Color
Wavelength
Frequency
2. Find the frequency for each color in the hydrogen spectrum. Do this by using the equation c =
fλ
3. Find the energy of light with a wavelength of 5000 Å (The unit "angstrom" is written Å and
hc
has a value of 10-10 meter.) Use the equation E 
(c = 3.00 x 108 m/s)

Calculate the energy for x-rays ( = 1 Å) and compare it to the 5000 Å light.
4. An incandescent bulb consists of a thin tungsten filament that is heated until it glows. If you
were to heat a different type of filament until it glowed, do you think it would produce a
continuous spectrum or a line spectrum? Why?
_______________________________
5. Which type of light is more energetic, red or violet? Calculate the energy of each photon to
support your answer.
57
G. LASERS
The laser is a unique source of e-m radiation produced by a process called stimulated emission.
It is used in many fields, including medicine, manufacturing, communications and construction.
Drilling through diamonds, measuring distances extremely accurately and repairing detached
retinas are just a few of the possible applications.
Properties of the laser
Radiation emitted by lasers is uniquely different from other sources of light such as incandescent
or fluorescent bulbs. Most light given off by these sources contain many different colors
(different wavelengths). Laser radiation consists of just one wavelength and thus laser light is
called monochromatic light. Most light sources radiate in all directions from the source and
thus their intensity is rapidly diminished as it moves farther away from the source. Laser light
leaves the source in a narrow beam that only spreads slowly as it moves away from the laser.
Thus, laser light possesses high directionality; it is called collimated light. Individual waves
from most light sources have no fixed phase relationship with each other. In laser light all the
waves are in phase with each other, a property called coherence. Light waves in phase with each
other interfere constructively and the resultant beam is thus stronger than a normal light beam.
Stimulated Emission
The word laser is an acronym which stands for Light Amplification by Stimulated Emission of
Radiation. It is this stimulated emission which gives the laser beam its unique properties.
The electrons about the nucleus of an atom seek their lowest, most stable energy state, or what is
called the ground state. When they are subject to outside forces, some of these electrons will
absorb energy sufficient to be excited to a higher energy level. A photon of the correct energy
can be absorbed by the electron, turning the photon energy into excitation energy of the electron.
The half-life of an excited electron is generally of the order of 10-8 sec. When it decays it will
return to its ground state, giving off a new photon of the same energy as the original photon.
Some elements have certain energy levels that are more stable, and an electron that reaches these
higher energy states can stay at this excited state much longer than usual before returning to the
ground state. The half-lives of these “meta stable states” can be on the order of 10-3 sec. This
allows a large population of excited electrons to exist. When a photon of the proper energy
interacts with the atom it causes the emission of many photons of the same energy. These
released photons all have the same wavelength, travel in the same direction and are in phase with
the incident photon. This stimulated emission produces a laser beam.
Parts of a laser
Three main components make up a laser: the lasing substance, the lasing cavity and the pump,
which is usually electrical or optical in nature. The lasing substance exhibits the properties
described above (it has a meta-stable state) and can be a solid, liquid or gas. It is contained the
lasing cavity, which is usually a hollow cylinder with mirrors at both ends. An energy source or
optical pump excites the electrons in the lasing substance to the meta-stable state and it gives off
laser light. The mirrors in the cavity reflect some of the light back and forth and allow the energy
to build up. The business end of the laser allows some of the light (about 1%) to escape the
cavity through a partially transparent mirror, as shown in Fig. II-23. Another gas, called a buffer
gas, is generally used to protect the lasing gas from de-excitation through collisions with the wall
of the lasing cavity.
58
100% reflective
mirror
monochromatic, coherent
stimulated emission
LASING CAVITY
99% reflective
mirror
emitted beam
Figure II-23 Laser schematic.
Laser Pumps
The process of supplying energy to the laser substance is called pumping. We call the electrical
or light source that delivers this energy a pump. Some lasers have electrical pumps such as an
AC or DC power source; this includes most gas lasers. Solid and liquid lasers have optical
pumps in which the laser substance absorbs the radiation from an intense light source.
Common Lasers
The helium-neon laser is one of the most common gas lasers and contains both helium and neon
gas. It is inexpensive and rugged. Besides being used in supermarkets to read bar codes, it is
used on construction sites and in laser pointers. It contains helium and neon gas in the cavity,
which is a glass tube with a mirror at each end. A DC power supply has wires into the cavity and
when it is turned on an electrical current passes through the tube. The gas mixture absorbs some
of this energy and converts it to light, which is reflected by the mirrors and amplified. Some of
the light is released through the partially reflecting mirror at one end and forms the output beam a bright red, pencil-like ray of light.
The carbon dioxide laser is another common laser which contains CO2 gas. This laser is used
in manufacturing to drill, weld and cut materials, including metal. The light produced is
ultraviolet, which is much higher in energy than the He-Ne laser, but invisible to the human eye,
so special precautions are necessary when using this laser.
The argon laser emits light in the blue and green portions of the visible spectrum. These are
used in laser light shows and in medicine, where they can remove skin blemishes or tattoos. The
green light is absorbed deeply enough by the skin to boil out the color of a tattoo, but not so deep
as to damage the skin. The argon laser is also used in eye surgery to repair detached retinas.
The neodymium-YAG laser uses solid yttrium aluminum garnet as the laser substance. It is
called "Nd-Yag" ("N-DEE-YAG"). This laser is optically pumped. An electrical current passes
through a tungsten-iodide pumping lamp to provide light energy. The Nd-YAG laser absorbs
this energy and releases laser light.
Dye Lasers use a liquid dye as the laser substance. In liquid dye lasers the wavelength can be
changed and thus these lasers are called tunable. Some can be tuned to the ultraviolet and
infrared wavelengths as well as visible light of various wavelengths.
59
Radiant power
We have said that laser light is very intense compared to ordinary light. To measure laser light
we need to know the radiant power, which is the amount of radiant energy delivered in a given
amount of time. Therefore, radiant power is measured in watts:
Radiant Power 
Radiant Energy
Time
J

  W
s

or
P
E
t
II-13
The shorter the time it takes to deliver the same amount of energy the greater the power, so time
is an important factor in the delivery of radiant energy.
Example II-15: Calculating radiant power
Two lasers like the one in Fig. II-23 each deliver 15 J of energy. Laser A does it in 150 μs while
laser B delivers it in 30 nanoseconds. What is the radiant power of each?
EA = EB = 15 J
tA = 150 s = 150 x 10-6 sec
tB = 30 ns = 30 x 10-9 sec
PA = ?
PB = ?
PA 
EA
15 J

 1 x 10 5 W
-6
t A 150 x 10 s
EB
15 J

 5 x 10 8 W
t B 30 x 10 -9 s
As you can see Laser B's power is 5000 times that of Laser A.
PB 
Power Density
Power density is the amount of power delivered per unit area. It is a better measure of laser
performance because it measures the intensity of the beam. Power density is also called
irradiance (). We’ll spend more time on this concept when we get into energy converters.
Power density is a key factor in the use of lasers. When the beam of a laser is focused through a
lens it allows the light energy to be concentrated. Power densities of over a million watts per cm2
can be achieved with the CO2 laser used for cutting metal. Lasers are especially good for cutting
as they do not heat up all of the metal, just the region being cut so the rest of the metal isn't
weakened or warped. Additionally, the cuts are clean and the laser doesn't wear out.
60
PROBLEM SET 5: LASERS (LAB 78)
1. Write an equation for irradiance, defining each term and listing the appropriate units.
2. Write an equation for radiant power, defining each term and listing the appropriate units.
3. A CO2 laser delivers a 40 watt beam which is focused on a target 0.1 cm in diameter.
Determine the power density.
4. An Nd-YAG laser is focused inside the eye to rupture an unwanted membrane. It delivers a
pulse of energy equal to 1 mJ. The pulse lasts for one nanosecond. The laser is focused on a
spot inside the eye 15 μm in diameter.
a) Find the power delivered by the laser pulse.
b) Find the power density of the laser pulse focused on the membrane.
61
5. A laser delivers 100,000 watts of power when 15 J of radiant energy are generated. Find the
time in seconds for the laser to deliver the energy.
6. Find the effect on the power of doubling the delivery time of the laser in problem #5 using
same energy.
7. A laser delivers 75 J of energy in one millisecond. Find the irradiance when the power is
focused by a lens onto a target area of 0.05 cm2.
8. A laser beam irradiance of 250 watts/cm2 is concentrated on a 0.2 cm2 area. Find the beam
power striking the target.
62
OVERVIEW:
LAB 78: LASER RADIATION
The word LASER is an acronym for Light Amplification by Stimulated Emission of Radiation.
The laser was proposed in the early 1900's by Albert Einstein but was only realized in the
1950's. The first laser, which actually generated radiation in the microwave region of the
electromagnetic radiation, was invented in the 1950's. It was called a maser, where the m
stands for microwave. The first visible laser was invented in the early 1960's. It generated
red light, much like the laser you will use. However, it was made using a ruby crystal. The
laser you will use contains a combination of helium and neon gases.
When the laser was invented, it was called "the invention without a use." Since those early days,
the laser has gained many uses. Lasers are commonly used in communications to transmit
data signals. They are used in industry for cutting, welding and measuring. The medical
profession uses lasers to cut tissue and to see inside the human body. The list of uses goes on
and on.
Laser radiation is a form of electromagnetic radiation. What makes it special? When you turn
on a light bulb, light spreads out in all directions. When you turn on a laser, only a pencilthin beam of light radiation comes out, and its divergence is very small. The beam is said to
be collimated. In fact, you will measure the nature of this beam in this experiment by
measuring its beam divergence. This is how much the pencil-thin beam spreads out as it
leaves the laser. The beam is also in phase; it is coherent. Another property of laser
radiation is that the color of the light emitted is a single color of the electromagnetic
spectrum; we say the beam is monochromatic. The helium-neon laser puts out a red beam
of light at a wavelength of 632.8 nm.
In this experiment, you will examine the color of the helium-neon laser beam. You will see how
the laser travels in space by observing how it passes through transparent materials.
You will also determine the beam’s spreading angle q, the angle at wish the beam expands as it
propagates. You will measure the near and far beam diameters (D1 and D2), and the
difference between the near and far distances (l) and calculate q via the equation given in the
Data Table. A diagram of the geometry of this experiment is shown below:
D1
(D2 – D1)


63
D2
LAB 78
OBJECTIVES:
LASER RADIATION
DATE______________
SKETCH:
OBSERVATIONS FOR PART A: FILTERS
Step A-9:
A-10:
__
A-11:
__
orange
yellow
purple
green
magenta
cyan
blue
red
Combinations of more than one filter:
Step B-2:
Step B-3:
DATA TABLE
Number of
Grids per
cm
(N)
BEAM DIVERGENCE
Beam Diameter
in Near
Position
D1
(# of grids)
Beam Diameter
in Far Position
D2
(# of grids)
Distance
Between
Readings
 (cm)
Beam Angle ()

D2  D1 

(rad)
64
LAB 78: ANALYSIS
1. When you add chalk to the water in part B, does this tend to focus or scatter the laser beam?
How does the light reach your eyes? Explain.
______________________________________
______________________________________
Understanding this, explain how you could make a laser beam visible in air when viewed from
the side (not reflected off something).
______________________________________
2. From parts A, B, and C of this lab, describe why laser light is different than that from an
incandescent or fluorescent bulb. List at least 2 differences.
_______________________________________
3. Five-Step: Suppose a reflector was set up on the moon, and you shined your laser at it. What
would the diameter of the reflector have to be to capture the entire beam? In other words,
what is the beam diameter at the moon? Assume the distance to the moon is 250,000 miles.
(250,000 miles = 400,000 km = 4.0 x 1010 cm). Use the results of step 3 of the Calculations
section. What is the beam diameter when the beam returns to earth?
The experiment above was performed by astronauts on the moon. A reflector was set up on
the moon and a laser was aimed at the reflector. The time for the laser to travel to the moon
and back was measured to determine the distance to the moon.
4. What color filter best absorbed the laser beam in Part A? Based on your answer, if you had to
pass as much laser light as possible through a blue filter, would you use a red laser (like the
helium-neon?) Explain.
_____
65
CHAPTER III: NUCLEAR RADIATION
A. LEARNING OBJECTIVES FOR CHAPTER III
1. Identify the three primary components of nuclear particle radiation: alpha particles, beta
particles and gamma rays.
2. Describe the relative hazards of alpha, beta and gamma radiation and their mitigation.
3. Define element, isotope, nuclide, atomic number and nucleon number.
4. Understand the basic structure and components of the atom.
5. Understand what is meant by fusion or fission and how each may occur.
6. Use Einstein's equation, E = mc2, in solving problems.
7. Solve nuclear equations involving beta decay, alpha emission, electron capture, position
emission and gamma emission.
8. Understand the process of exponential decay.
9. Understand the units by which radiation is measured.
B. INTRODUCTION
The quest to understand what matter is has made significant strides in the last century. We know
that matter is made up of atoms. Nearly 100 years ago Rutherford discovered the atomic nucleus,
the dense center of the atom with a diameter 100,000 times smaller than the complete atom. It
contained protons (m ≈ 1.673 x 10-27 kg) with a positive charge equal to the negative charge of
the much less massive electrons (m ≈ 9.109 x 10-31 kg) that surrounded it. Shortly thereafter
Bohr devised a planetary model of the atom that explained Rutherford’s findings. He envisioned
negative electrons orbiting the nucleus much as the planets orbit the sun, but with a twist. The
electrons were constrained to certain orbits; they could never be between these orbits and they
could never cross the boundaries in transit. “Jumping” to a higher orbit meant gaining the exact
right amount of energy to stop being at one orbit and start being at another. Such is the strange
world of quantum mechanics.
In the 1920’s the planetary model was modified. Electrons had wave properties as well as
particle attributes, and they didn’t orbit about the nucleus like planets. Their wave characteristics
were more important than previously understood. Their orbital radius was determined by the size
of the electrons’ associated wavelength; the radius was the radius of a standing wave, an electron
standing wave, wrapped around in a circle. Higher orbits were merely higher harmonics of a
standing wave loop. Just as it takes more energy to form higher harmonic standing waves with a
slinky, electrons with higher energy formed higher harmonic standing waves at greater distances
from the nucleus. Electron “orbits” were more properly called electron “energy levels”.
In the 1930’s the neutron was discovered. Neutrons are particles just slightly more massive than
the proton (~1.675 x 10-27 kg), but with no charge. They reside in the nucleus and seem to shield
the protons’ positive e-m charges from each other, stabilizing the nucleus. Changes in the
nuclear structure of the atom involve several millions of times more energy than changes in the
electron structure of an atom, such as those occurring with the emission of photons or in the
formation of various chemical bonds between atoms. This is why a 1 kg atomic bomb releases
so much more energy than 1 kg of TNT.
66
In this chapter we take a brief peek into the structure of the atom and the processes governing
nuclear radiation.
C. STRUCTURE OF THE ATOM
The atom consists of a nucleus and a cloud of electrons surrounding the nucleus. The nucleus is
held together by forces much stronger than those that bind electrons to that atom. Electrons can
be released from the atom relatively easily, while it is extremely difficult to break apart the
nucleus.
Each of the different atomic elements contains a unique number of protons and a varying number
of neutrons. An electrically neutral atom contains equal numbers of electrons and protons, and
the positive and negative charges cancel out. If the number of electrons is different from the
number of protons, the atom is called an ion; positive ions (cations) have more protons than
electrons and carry a positive electrical charge. Negative ions (anions) have fewer protons than
electrons and carry a negative electrical charge.
The simplest atomic structure is the element hydrogen. Its nucleus consists of a proton with a
single electron in orbit about the nucleus. The mass of the nucleus is 1.673 x 10-27 kg, while the
mass of the electron is 9.109 x 10-31 kg. The mass of the proton is 1836 times greater than that of
the electron. As you can see, most of the mass of the atom resides in the nucleus; this is true for
all elements. A small fraction of hydrogen atoms have a neutron in the nucleus, and an even
smaller fraction of hydrogen nuclei contain two neutrons. All three of these atoms are called
hydrogen because they contain one proton. The number of neutrons in the nucleus tells us what
isotope of hydrogen we’re dealing with. The isotope of hydrogen that contains one neutron is
called deuterium. The isotope containing two neutrons is called tritium. Ordinary hydrogen
and deuterium are both stable and occur in nature. Tritium is highly radioactive and will
gradually decay to become an isotope of the helium atom. Tritium is formed by cosmic radiation
from space and in the decay of nuclear fuel. At any given time there are only about 2 kg of
naturally occurring tritium on earth, most of which is in the oceans of the world. All of the other
elements contain more protons than hydrogen – the number of protons in the nucleus defines the
element. Each element has different isotopes, depending on the number of neutrons in its
nucleus.
Protons and neutrons are called nucleons because they reside in the nucleus of the atom. The
number of protons in the nucleus is the atomic number (symbol Z) of the element. For all
elements except the very lightest, the number of neutrons (N) in the nucleus equals or exceeds
the number of protons. The total number of nucleons (protons and neutrons) in an atom is its
nucleon number, also known as the mass number (symbol A). The conventional symbols for
nuclear atoms or nuclides all follow the same pattern ZXA, (or AZ X ), where X is the chemical
symbol for the element. Ordinary hydrogen would be designated 1H1, while deuterium would be
2
4
1H . The most common isotope of helium is 2He . Isotopes of elements heavier than hydrogen
are also listed by their name, followed by their nucleon number, because if you know the element
name, you know how many protons it has. 2He4, for example, is commonly known as “helium4”. The compositions of common isotopes of the three lightest elements, hydrogen, helium and
lithium, are shown in Fig. III-1.
67
1
1H
4
2He
3Li
6
Figure III-1. Composition of common isotopes of hydrogen (1H1), helium (helium-4) and lithium
(lithium-6). The black circles are protons and the white circles are neutrons.
Example III-1: Understanding the Periodic Table
Name the chemical element, Z, A, and N for 1H2, 29Cu64 and 92U235.
2
1H :
Hydrogen, Z = 1, A = 2, N = 1
64
Cu
: copper, Z = 29, A = 64 and N = 35
29
235
92U : Uranium, Z = 92, A = 235 and N = 143
The chemical properties of an element depend primarily upon the distribution of the electrons
about the nucleus and not the structure of the nuclei, aside from the number of protons.
Therefore, the isotopes of an element will have nearly the same chemical properties. The
physical properties of an element depend upon the nuclear structure, so they can vary from one
isotope to another. For example, the slight difference in weight between isotopes of the same
element allows us to physically separate them from each other.
Periodic Table of the Elements
All of the elements have been arranged in what is called the periodic table of the elements. In this
table, shown in Fig. III-2, you can find the atomic weight, the atomic symbol, the atomic number,
whether the element acts as a metal or as a non-metal, and additional information, depending on
the table. Atomic numbers from 1 to 109 (and higher!) have been identified, but only those with
atomic numbers below 94 occur naturally. The nucleon number (or mass number, A) is usually
listed as an average of its isotopes, weighted according to each isotope’s relative abundance.
The format of this periodic table is designed to highlight chemical similarities between the
elements. Electrons in an atom are grouped in shells, much like the layers of an onion.
Generally, only the electrons in the outermost shell react chemically with other elements – these
are known as the valence electrons. The innermost shell holds two electrons, the next two shells
holds eight. The next three contain 18 electrons. Each shell is listed as a row in the chemical
table. Each column of the table contains the same number of valence electrons. The most
chemically stable configuration for an atom or molecule is to have its electron shells completely
filled and still be neutral.
Look at the column of elements on the very right side of the table, with helium (He) at the top.
Helium has two electrons, which fills its electron shell. The next element down in the column is
neon (Ne). It has an inner electron shell that holds two electrons, like helium, and an outer
(valence) shell that holds eight. Neon has ten protons, so the neutral atom has ten electrons,
which fill the shells. Helium, neon, argon (Ar), krypton (Kr) and xenon (Xe) all have their
68
electron shells filled. They are unlikely to share their electrons with anybody else. The elements
in this column are called noble gases, because they like to be alone. When all the available
electrons positions are
filled in a shell (complete shell), the atom is very stable and rarely interacts chemically with other
atoms. It is very rare to find a chemical reaction involving a noble gas.
69
LIGHT
METALS
PERIODIC TABLE OF THE ELEMENTS
IA
VIIIA
Hydrogen
Helium
1.0000
4.003
In the periodic table the elements are arranged in order of increasing atomic number.
NON METALS
He
Vertical columns headed by Roman. Numerals are called Groups. A horizontal
sequence is called a Period. The most active elements are at the top right and
left of the table. The staggered line (Groups IIIA and VIIA) roughly separates IIIA
IVA
VA VIA VIIA
2
1 IIA bottom
metallic from non-metallic elements.
Lithium Beryllium Group IA includes hydrogen and the alkali
Carbon
Nitrogen Oxygen
Fluorine
Neon
Groups- Elements within a group have Boron
metals.
similar properties and contain the
6.939 9.012 Group VIIA includes the halogens.
10.811 12.01115 14.007 15.999
18.998 20.183
same number of electrons in their
outside energy shell.
elements intervening between groups IIA
Li Be The
B
C
N O
F Ne
and IIIA are called transition elements.
elements intervening between groups IIA Periods- In a given period the
3
4 The
5
6
7
8
9 10
and IIIA are called transition elements.
properties of the elements gradually
pass from a strong metallic to a strong
nonSodium Magnesium The elements intervening between groups IIA metallic nature, with the last number of Aluminum
Silicon Phosphate Sulfur
Chlorine
Argon
and IIIA are called transition elements.
a period being An inert gas.
H
22.990 24.312
Na Mg
11
Short vertical columns without Roman
numeral headings are called sub-groups.
12
Potassium Calcium
Scandium
Titanium
Vanadium Chromium Manganese
Cobalt
Al
Si
13
14
Nickel
Copper
Zinc
Gallium
Germanium
58.71
63.54
65.37
69.72
Cu Zn
29 30
Ga
31
72.59
74.922 78.96
79.909
83.80
Ge
32
As Se
33 34
Br
35
Kr
36
K
19
Ca
20
Sc
21
Ti
22
V
23
Cr
24
Rubidium Strontium
Yttrium
Zirconium
115.17
87.62
88.905
91.22
92.906
95.94
99
Rb
37
Sr
38
Y
39
Zr
40
Nb
41
Mo
42
Tc
43
Cesium
Barium
Hafnium
Tantalum
Tungsten
Rhenium
Osmium
Iridium
Platinum
Mercury
Thallium
Lead
132.90 137.34
178.49
180.95
183.85
186.21
190.2
192.2
195.09
196.97 200.59
204.37
207.19
Ta
73
W
74
Re
75
Pt
78
Au Hg
79 80
Tl
81
Pb
82
Ru Rh Pd
44 45 46
Os Ir
76 77
Silver
Indium
Tin
Antimony Tellurium
Iodine
Xenon
107.87 112.40
114.82
118.69
121.75 127.60
126.90
131.30
Ag Cd
47 48
In
49
Sn
50
I
53
Xe
54
Gold
Cadmium
16
Ar
18
51.996
106.4
15
Cl
krypton
50.942
101.07 102.91
S
39.948
17
47.90
Niobium Molybdenum Technetium Ruthenium Rhodium Palladium
P
35.453
Bromine
44.956
Mn Fe Co Ni
25 26 27 28
30.974 32.064
Arsenic Selenium
40.08
Ba 57-71 Hf
56
72
55.847 58.933
28.086
39.102
Cs
55
54.938
Iron
26.981
Sb
51
Te
52
Bismuth Polonium
Astatine
Radon
200.98
(210)
(210)
(222)
Bi Po
83 84
At
85
Rn
86
Francium
Radium
22.3
(226)
Fr
87
Ra 89-103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118
88
Lanthanum Cerium
138.91 140.12
Praseodymium Neodymium Promethium Samarium
144.24
Ce
58
Pr
59
Nd
60
Actinium
Thorium
Protactinium
Uranium
227
232.04
(231)
238.03
(237)
(242)
(243)
Pa
91
U
92
Np
93
Pu
94
Am Cm Bk Cf
95 96 97 98
La
57
Ac Th
89 90
(147)
150.35
Europium Gadolinium Terbium Dysprosium Holmium
140.91
Pm Sm
61 62
151.96
157.25 158.92 162.50
Eu Gd Tb Dy
63 64 65 66
Neptunium Plutonium Americium
Erbium
Thulium
Ytterbium
Lutetium
164.93 167.26
168.93
173.04
174.97
Ho Er
67 68
Tm
69
Yb
70
Lu
71
Curium Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium
(247)
(249)
(251)
(254)
(253)
(256)
(254)
(257)
Es Fm Md No Lr
99 100 101 102 103
Figure III-2. The Periodic Table of the Elements (Mendeleev version).
70
The other elements tend to combine with each other in ways that makes the combination inert.
Look at the first column of elements on the left. Hydrogen, at the top, contains one valence
electron in its outermost shell which, in this case, is the innermost shell. Directly below
hydrogen are lithium (Li), sodium (Na) and potassium (K), in that order. Each of these elements
has one valence electron. All of these atoms have an affinity for a chemical reaction that would
fill their electron shells. Sodium needs seven more electrons to fill its shell. Meanwhile,
chlorine in the second column from the right, only needs one more electron to fill its shell. If
chlorine could share its seven valence electrons while sodium shares its one valence electron,
together sodium and chlorine could share their valence electrons and have a full shell of eight,
and be chemically stable. That is why sodium chloride (salt) is so common; together the two
atoms can fill their electron shells. Potassium chloride would also work, because potassium has
one valence electron, just like sodium. Potassium chloride is commonly used as a salt substitute.
As another example, check out oxygen (O). Oxygen is two valence electrons short of a full shell.
When it combines with two hydrogen atoms, it shares an electron with each hydrogen atom and
they each share their electron with oxygen. Together, H2O (water) has eight valence electrons – a
full shell. Some animals located near thermal vents at the bottom of the ocean are able to breathe
H2S instead of H2O, because they react chemically in similar ways. Note that sulfur (S) is in the
same column as oxygen. Copper (Cu), silver (Ag) and gold (Au) have very similar electrical and
chemical properties. See if you can find their relative positions in the periodic table.
Atoms with their valence shells less than half full tend to share their few valence electrons more
easily and are typically good conductors. Atoms with more than half of their shells filled hang
on to their valence electrons more tightly and are typically good insulators. Electrons with
exactly half of their shell filled are in between, and are called semi-conductors. The semiconductor column includes carbon (C), silicon (Si) and germanium (Ge). The table of elements
contains far more information than we have provided here, but we must press on to the center of
the atom, to study the processes that govern its nucleus.
71
PROBLEM SET 6: ATOMIC STRUCTURE
1. Find the atomic number, the nucleon (or mass) number and the number of neutrons for the
following elemental species:
a) 13Al27
b) 38Sr88
c) 4Be9
d) 79Au197
e) 6C12
2. List the elements with the chemical properties similar to a) helium (He) and b) boron (B)
72
D. NUCLEAR INTERACTIONS
Opposite electrical charges attract and like charges repel. Positively charged protons are attracted
to negatively charged electrons, but electrons repel each other and protons repel each other. Yet
the nucleus contains positively charged protons, huddled together in a space tens to hundreds of
thousands of times smaller than the size of the total atom. The force that allows them to do this
is called the strong force. The strong force is much stronger than the electromagnetic force over
very short distances (on the order of 10-15 m), but it dissipates much faster than the
electromagnetic force. It takes enormous energies to get two protons close enough to each other
for the strong force to take over, but once it does those protons are actually attracted to each
other. The small interaction distance, however, works against the stability of the nucleus. As the
protons get further from each other, the strong force becomes weaker and the electromagnetic
(repulsive) force becomes more important. Protons on either side of a nucleus can feel more of
the electromagnetic force than the strong force. That’s where neutrons come in. Somehow these
uncharged particles shield the protons from each others repulsive forces, allowing them to pack
together in a more stable manner. No multi-proton nucleus, and that means anything other than
hydrogen, can exist without neutrons to keep the protons together. Heavier elements, with more
protons and larger diameters, require more and more neutrons to keep them together. While
helium (two protons) commonly has four nucleons (two protons and two neutrons), uranium-238
(92U238) has 146 neutrons holding its 92 protons together.
The graph in Fig. III-3 plots protons (Z) vs. neutrons (N) for stable nuclei. All the stable nuclei
with more than 20 protons contain more neutrons than protons, and the discrepancy increases
with the number of protons. Below 20 protons the number of protons is approximately equal to
the number of neutrons. Evidently neutrons are an extremely important packing material in the
atomic nucleus.
Figure III-3. Graph of number of protons vs. number of neutrons for stable atomic nuclei.
73
Nuclear Synthesis – Fusion and Fission
It appears that the visible matter in the universe 13.7 billion years ago consisted of mostly
hydrogen and some helium, with a very small smattering of lithium. All of the heavier elements
have formed in the intervening years. The building up of nuclei with larger atomic numbers from
smaller elements is called nuclear fusion. Fusing protons and other nuclei together requires
high temperatures and pressures to overcome the electromagnetic force that causes protons to
repel each other. These conditions are found at the core of stars. The presence of carbon-based
life forms breathing oxygen and nitrogen on a planet made of rock and metal is evidence that we
are the result of stars that lived and died long before we were formed. Hydrogen nuclei in the
core of our sun are constantly being forced together to produce helium nuclei. This process has
been going on for nearly five billion years, and it appears it will continue for another five billion
years. More massive stars, like the ones that made the atoms we’re composed of, have denser,
hotter cores and fuse together heavier elements like carbon, oxygen, and neon – all the way up to
iron (Fe). Elements with atomic numbers greater than iron are formed in the shock waves
produced when still larger stars die in an explosive demise known as a supernova. The gold and
silver in your teeth were formed in a supernova explosion.
The curve of binding energy
Chemical reactions occur in the direction that produces products with less energy than their
reactants. Automobile exhaust, for example, contains less energy than the gasoline and oxygen
that combined to form it. The energy was released and partially used to make the car move.
Nuclear processes work in the same way: they occur in a direction such that the products are in a
lower energy state than the reactants and, in the process, they give off energy. In our sun, protons
fuse together to form deuterium. This must seem strange to you, since deuterium is composed of
a proton and a neutron, but there is more physics going on here than we’re going to cover in this
text. We will say that in some processes, the proton can absorb an electron and become a
neutron – this is called electron absorption. Anyway, we form deuterium (1H2) in the sun. Two
deuterium nuclei can then combine to form helium-4 (2He4), as in Fig. III-4.
+
2
1H
+
2
1H
4
2He

Figure III-4. Nuclear fusion of two deuterium nuclei to form helium-4 and a gamma ray photon.
In the process, energy is released in the form of high energy photons in the gamma ray region of
the electromagnetic spectrum. These gamma ray photons (denoted by “”, the Greek letter
“gamma”) take hundreds of thousands of years to bounce their way from the core of the sun to
the surface, losing energy to their surroundings as they go. By the time they reach the surface,
most have lost so much energy that they are in the visible spectrum. The subsequent trip from
the sun’s surface to the Earth takes a little over eight minutes. Fusion is the energy production
process that provides the sunlight that sustains life on Earth.
We can see where the energy in a fusion reaction comes from by adding up the masses of the
74
reactants and the products. The mass of a proton (to four significant figures) is 1.673 x 10-27 kg
and the mass of a neutron is 1.675 x 10-27 kg. A helium nucleus is comprised of 2 protons and
two neutrons. The sum of the individual components is
msum = 2mproton + 2mneutron = 2(1.673 x 10-27 kg) +2(1.675 x 10-27 kg) = 6.696 x 10-27 kg
The mass of a helium nucleus however, is 6.645 x 10-27 kg, three quarters of a percent less than
the sum of its components! The product (2He4) has less mass than the sum of the particles
that produced it. The missing mass has been converted into energy. About a hundred years
ago, Albert Einstein discovered that mass is just another form of energy, sort of like “frozen”
energy. He described the relation between energy (E) and mass (m) with the following equation:
E = mc2
III-1
where c is the speed of light in vacuum, 3 x 108 m/s.
Example III-2: Energy conversion during the production of helium nuclei.
In the above example, the mass of the helium nucleus is 5.1 x 10-29 kg less than the sum of the
protons and neutrons of which it is composed. How much energy is released in this process?
m = 5.1 x 10-29 kg
E=?
E = mc2 = (5.1 x 10-29 kg)(3.0 x 108 m/s)2 = 4.59 x 10-12 J
This doesn’t seem like much energy – 5 trillionths of a joule – but nuclei aren’t very big. If you
had enough of them together you can release quite a bit of energy.
Example III-3. Energy stored in a liter of beer
If you could convert all of the mass in one liter of beer (= 1 kg) into energy, how much energy
would you have?
Normally I hate to waste a beer, but this is in the name of science, so I guess it’s okay. Don’t
count on the glass being there afterward either. Incidentally, it doesn’t matter if it’s beer or lead
– all that matters is the mass.
m = 1 kg
E=?
E = mc2 = (1 kg)(3 x 108 m/s)2 = 9 x 1016 J
90 million billion joules sounds much more impressive, but let’s convert it to the equivalent
energy produced by detonating TNT. One ton of TNT releases 4.18 x 109 J of chemical energy,
so
75


 1 ton TNT 
7
E  9 x 1016 J 
  2.15 x 10 tons TNT
9
 4.18 x 10 J 
For a one ton pick-up truck, that’s over 20 million pick-up trucks full of TNT. This is just
further proof of the potential dangers of alcoholic beverages.
Binding Energy
Stable nuclei always have less mass than the combined masses of their constituent particles.
This difference in mass is called the binding energy. Fig. III-5 is a plot of the binding energy
per nucleon (the protons and neutrons) vs. nucleon number for stable nuclei.
As we start from hydrogen (A=1), the binding energy increases fairly smoothly, except for the
peak at 2He4. It peaks at A=56, which is iron-56 (or Fe-56). Beyond this point the binding
energy per nucleon decreases with increasing nucleon number. Nuclear processes move to lower
energy by moving in the direction that increases their binding energy per nucleon. To the left of
iron-56, nuclei undergo fusion, that is, they fuse together to become heavier nuclei and release
energy in the process. To the right of iron-56, nuclei break up to release energy. This process of
breaking apart to release energy is fission. It is this type of nuclear reaction that powers every
nuclear reactor in the world today. The most stable particle is iron-56; lighter nuclei undergo
fusion until they become iron-56 and heavier particles undergo fission until they get down to
iron-56.
Scientists and engineers have been working for years to produce a fusion reactor that produces
more power than is required to run it, but they have been unsuccessful to date. Matching the
conditions at the core of the sun is obviously no easy task, but the potential rewards of success
are great. Fusion reactions are seven times more efficient at producing energy than fission
reactions. Furthermore, fusion reactions produce no radioactive waste, which is a topic of great
concern for the current nuclear power industry.
But where did the elements heavier than iron-56 come from? As stated before, the only known
method of producing these elements is from the sudden compression and heating created in the
shock wave of a supernova, the explosive demise of a star much larger than our own.
Figure III-5. The curve of binding energy: binding energy per nucleon vs. the number of
76
nucleons (protons and neutrons) in the nucleus. The peak occurs at iron-56.
To summarize, fusion is the merging of smaller nuclei to form larger nuclei and fission is the
process by which larger nuclei break up into smaller nuclei.
Radioactive decay
Not all nuclei are stable. This was discovered by the research of Becquerel, the Curies and many
others. Some nuclei spontaneously transform themselves into other nuclear species with the
emission of radiation from the nucleus. Such nuclei are said to be radioactive. Some types of
nuclei can, on average, last for billions of years while others last billionths of a second. The
process by which nuclei destabilize and emit particles and e-m waves is called radioactive
decay.
Unstable nuclei emit three main types of radiation from the nucleus: alpha particles, beta
particles and gamma rays. Alpha particles () are the nuclei of the helium atom, 2He4. An
unstable nucleus becomes lighter by emitting an alpha particle, a heavy type of radiation
containing two protons and two neutrons. Beta particles ( or -1e0) are electrons created in and
emitted at high speed from unstable nuclei. Beta decay occurs when a neutron spontaneously
changes into a proton and an electron. This can happen inside or outside the nucleus. A gamma
ray () is a form of electromagnetic radiation given off by the nucleus in a decay process.
Gamma rays are at the high energy, high frequency, short wavelength end of the e-m spectrum.
These and a few other types of radioactive decay are described below. It will be easier to
describe the processes using the symbols for each particle in the form of a nuclear equation.
Don’t be scared by the phrase “nuclear equation”. It’s just bean counting. The reactants are on
the left side of an arrow showing the direction of the reaction and the products are on the right.
The sum of the atomic numbers (subscripts) of the particles on the left side must equal the sum
of those on the right side, and the sum of the nucleon numbers (superscripts) on the left side must
equal the sum of those on the right side. If you know the atomic number, you can use the
Periodic Table to find the chemical symbol of the element. You should be aware of a few extra
symbols that aren’t in the Periodic Table. The electron is denoted by -1e0, so it has a negative
atomic number and zero nucleon number (no nucleons). The neutron is denoted by 0n1. We’ll
discuss one anti-matter reaction, the positron (1e0). It’s just like the electron except it has a
positive charge. It is said to be the anti-matter counterpart of the electron. It’s also called the
anti-electron. When matter meets its anti-matter counterpart, the two particles annihilate, that is,
their mass is converted into pure energy. Every matter particle has an anti-matter counterpart,
but it is one of the great mysteries of the universe why there is so much more matter than antimatter. Anyway, here are the radioactive decay reactions we will study:
1. Beta Decay
If a nucleus has too many neutrons for stability relative to the number of protons, it is radioactive
and will decay. If a neutron turns into a proton and an electron, there will be 1 less neutron and 1
more proton. The electron gets kicked out of the nucleus – they’re not welcome in there:
Beta Decay
n1 → 1H1 + -1e0 + 
0
III-2
The electron leaves the nucleus at high speed, but since there is excess energy remaining, a
77
gamma ray is given off. Sometimes more than one beta particle is required for the nucleus to
reach a stable state. Note that the sum of the atomic numbers on the left (0) is equal to the sum
of the atomic numbers on the right, and the sum of the nucleon numbers on the left (1) is equal
to the sum of the nucleon numbers on the right.
1
1H
1
0n

0
-1e
Figure III-6. A beta decay cartoon depicting the spontaneous decay of a neutron into a proton, an
electron and a gamma ray.
2. Positron Emission
Should the nucleus have too few neutrons, the inverse reaction takes place and a proton becomes
a neutron by emitting a positron:
Positron Emission
1
H1 → 0n1 + +1e0 + 
III-3
A positively charged electron (positron) is emitted from the nucleus along with a gamma ray, as
shown in Fig. III-7.
1
0n
1
1H

0
+1e
Figure III-7. The emission of a positron from a proton converts it into an electron.
3. Electron Capture
This is a process where one of the innermost orbiting electrons is absorbed into the nucleus by a
proton, which turns into a neutron:
Electron Capture
H1 + -1e0 →
1
78
n1
0
III-4
During this process an x-ray is given off by an orbiting electron as it replaces the captured
electron. It is this x-ray that tells us an electron has been captured. See Fig. III-8.
x-ray
0
-1e
0n
1
1
1H
Figure III-8. Electron capture is the reverse of beta decay.
4. Alpha Decay
Helium-4 nuclei, or alpha (“a”) particles, are very stable. Often a nucleus will give up energy by
emitting an alpha particle. Such is the case for polonium-210:
Alpha Decay
84Po
210
→
82Pb
206
+2He4
III-5
206
82Pb
210
84Po
4
2He
Figure III-9. Alpha decay of polonium-210 produces lead-206 and an alpha particle.
5. Gamma Ray Emission
Sometimes, generally in the midst of a series of reactions, a nucleus will reach an excited state
(much like an atom with an electron moving to a higher energy level) and it will spontaneously
emit a gamma ray. The resulting nucleus is the same as before, but with less energy. As an
example,
Gamma Ray Emission
131*
54Xe
→
79
131
54Xe
+
III-6
In summary, negative beta decay increases Z by 1 and decreases N by 1, positive beta decay and
electron capture reduce Z by 1 and increase N by 1 and alpha decay reduces both Z and N by 2.
The table below shows each of these processes and gives an example of each of them.
Table of Decay Processes
Decay Type
Nuclear Reaction
Example
Alpha
A
ZX
→ Z-2YA-4 + 2He4
238
92U
Electron emission
A
ZX
→ Z+1YA + e-
6C
Positron emission
A
ZX
→ Z-1YA + e+
29Cu
Electron capture
A
ZX
+ e- → Z-1YA + γ
Gamma ray emission
A*
ZX
64
29Cu
87*
38Sr
Example III-4: Write the equation for the alpha decay of
83Bi
208
→ 7N14 + -1e0
14
64
→ ZXA + γ
83Bi
208
→ 90Th234 + 2He4
→ 28Ni64 + +1e0
+ -1e0 → 28Ni64 + γ
→ 38Sr87 + γ
and show the products.
→ 81Tl204 + 2He4 (alpha)
Z decreases by two from 83 to 81 which is thallium (Tl) and A decreases by 4 (2 protons and 2
neutrons) from 208 to 204.
Example III-5: What does a nucleus of 29Cu64 become when it captures a bound electron?
29Cu
64
+ -1e0 → 28Ni64
The absorbed electron turns one proton into a neutron, reducing Z by 1 and increasing N by 1, so
that A remains the same.
Example III-6: A 35Br80 nucleus can decay by beta decay, positron emission or by electron
capture. Write the equation for each of these processes.
Beta decay:
80
35Br
→ 36Kr80 + -1e0 + γ
Positron decay:
80
35Br
→ 34Se80 + +1e0 + γ
Electron capture:
80
35Br
+ -1e0 → 34Se80
80
PROBLEM SET 7: NUCLEAR RADIATON
1. Identify the chemical element, Z, A, and N for 11Na24 and 82Pb210.
2. The ink used to print the words: Nuclear Radiation has a mass of about 3 x 10-6 g. If this
mass were all converted to energy, what energy would we have?
3. Radium undergoes spontaneous decay into helium and radon. Why is radium regarded as an
element rather than a chemical compound of helium and radon?
81
Lab 77 B: Calculating Radioactive Decay Products
Objectives
Simulate the process by which Uranium-238 eventually decays to Lead-206
Overview
Each element is defined by the number of protons in its nucleus. Hydrogen, for example, always
contains one proton, while carbon always contains six. Each element also has different isotopes;
that is, they can contain different numbers of neutrons. Hydrogen nuclei usually contain one
proton and no neutrons (1H1), but occasionally there may be a neutron in the nucleus (1H2), and
on still rarer occasions, we may find two neutrons (1H3), but there is always one proton. Protons
and neutrons are the only inhabitants of atomic nuclei, which is why both they are also known as
nucleons. Neutrons seem to assist in keeping the positively charged protons together in the
tightly bound nucleus. More protons in a nucleus require even more neutrons. The most
common isotope of Carbon, Carbon-12 (6C12), has 6 protons and 6 neutrons, while Uranium-238
(92U238) has 92 protons and 146 neutrons (more than one and a half times as many neutrons as
protons).
The different isotopes of each element have different levels of stability. Some have half-lives of
billions of years, while others have half-lives on the order of a few millionths of a second. They
decay via radioactive processes, a few of which we will study today. Nuclei can absorb and
emit electrons, neutrons, protons, alpha particles (helium nuclei), and positrons (“positive
electrons”). They also emit gamma radiation, which removes energy from the nucleus without
changing the isotopic structure. (Incidentally, the beta decay process also emits a particle known
as a neutrino, an extremely small mass particle with no charge. Although they are very
important in the make-up of the universe, they don’t interact with your body, so we won’t be
discussing them.) The decay process conserves the atomic number (the total number of
protons) and the nucleon (or mass) number (the total number of protons and neutrons, that is,
the total number of nucleons).
In this lab, we’ll simulate the decay of Uranium-238 (92U238) into Lead-206 (82Pb206), a common
process on our planet. Radon-222 (86Rn222) is an intermediate stage of the decay process - you
may have heard about homes being checked for Radon. As you’ll find out, inhaled Radon winds
up as lead in your lungs, and along the way it emits alpha particles (helium nuclei), beta particles
(electrons) and gamma rays, none of which are particularly good for your health. Unfortunately,
protons and neutrons are difficult to work with in a hands-on experiment, so our protons will be
black beans, our neutrons will be white beans, and our nucleus will be a glass jar. We’re going
to be nuclear bean counters.
Procedure
Empty the jar’s contents onto the desk and count the beans. This should be a Uranium-238
nucleus, so the jar should contain 92 protons (black beans) and 146 neutrons (white beans). Put
them in groups of ten while counting to ensure accuracy. If you need more beans, tell the lab
82
tech or your instructor. After you’ve made sure you have the right number of nucleons, put them
back in the nucleus (jar). Keep in mind that these decay processes have widely varying halflives. The half-life of U-238, for example, is 4.5 billion years, while the half-life for Polonium214 is 0.00015 seconds
1) First Reaction:
The first reaction for U-238 is alpha decay, which is the loss of a helium nucleus (also known as
an alpha particle) from the “parent” nucleus. The half-life of U-238 is 4.46 billion years. The
half-lives of each product are given in parentheses. A helium nucleus contains 2 protons and 2
neutrons. Remove one helium nucleus (alpha particle) from the jar. What isotope is left? The
nuclear reaction equation will tell you, because you know that atomic number and nucleon
number are conserved:
92U
238
 90Th234 + 2He4
Isotope Name: Thorium-234
(24.1 days)
The nucleus now has 90 protons (= 92 – 2) and 144 neutrons (= 146 – 2). Looking at the
periodic table in your text, you’ll find that the isotope must be Thorium-234.
2) Second Reaction:
Thorium-234 decays via beta decay, that is, by ejecting an electron from the nucleus. The beta
decay process also emits a gamma ray. What actually happens is that a neutron is converted to a
proton and an electron, and the electron is ejected. You can simulate this by removing a neutron
(white bean) and inserting a proton (black bean). The reaction equation tells us what’s left. An
electron is listed as having an atomic number of –1 and a nucleon number of zero. Making sure
that atomic number and nucleon number are conserved, we can figure out what is left after beta
decay:
234
90Th
 91Pa234 + -1e0 +
Isotope Name: Proactinium-234
(1.17
min.)
The product of beta decay of Thorium-234 is Proactinium-234; 91 protons and 143 neutrons (one
less neutron than Thorium-234, but one more proton).
The Rest of the Reactions
For the remaining reactions, we’ll tell you what particle is lost and you’ll adjust the nucleons in
your jar accordingly. Remember, alpha decay means you remove two protons and two neutrons,
beta decay means you replace a neutron with a proton, and gamma decay doesn’t change the
number of protons and neutrons. Fill in the blanks in the equation for each step and include the
name of the isotope, as we did in steps 1 and 2. In the end you should wind up with a jar of
206
(also known as Lead-206) a stable (but chemically toxic) isotope of lead. When you’re
82Pb
finished, count the beans to check your results.
3) Beta Decay:
234
91Pa
 + ________ + -1e0 + 
Isotope Name: _________________ (245,000 yrs)
4) Alpha Decay:
_______  _______ + 2He4
Isotope Name: _________________ (75,400 yrs)
83
5) Alpha Decay:
_______  _______ + 2He4 + 
Isotope Name: _________________
(1,600 yrs)
Isotope Name: _________________
(3.82 days)
Isotope Name: _________________
(3.11 min.)
Isotope Name: _________________
(26.8 min.)
Isotope Name: _________________
(19.9 min.)
6) Alpha Decay:
_______  _______ + 2He4 + 
7) Alpha Decay:
_______  _______ + 2He4
8) Alpha Decay:
_______  _______ + 2He4
9) Beta Decay:
_______  _______ + -1e0 + 
10) Beta Decay: (This step and the next are often reversed)
_______  _______ + -1e0 + 
Isotope Name: _________________ (1.63x10-4 sec)
11) Alpha Decay:
_______  _______ + 2He4
Isotope Name: _________________ (22.3 yrs)
12) Beta Decay:
_______  _______ + -1e0
Isotope Name: _________________ (5.01 days)
13) Beta Decay:
_______  _______ + -1e0
Isotope Name: _________________ (138 days)
14) Alpha Decay:
_______  _______ + 2He4
Isotope Name: _________________ (relatively
stable)
15) Count your beans to check your results.
84
LAB 77B: ANALYSIS
1) How many alpha particles were emitted in the 14 decay processes?
2) How many electrons were emitted?
3) How many gamma ray photons were emitted?
4) If a person inhales Radon-222 and it stays in their lungs until it becomes Lead-206, how
many alpha particles are emitted into your lungs for each Radon-222 atom?
5) How many electrons are emitted into your lungs?
6) How many gamma ray photons?
85
E. HALF-LIFE OF RADIOACTIVE ISOTOPES
If radioactive decay was a linear process, the number of decays per year would be constant, but
that’s not how it works. The time period for the decay of an individual radioactive atom is a
random process, but the rate at which a large collection of atoms decays can be precisely
determined. Statistically, radioactive elements decay in an exponential manner. A typical
exponential decay curve is shown in the graph in Fig. III-10, with the number of radioactive
particles plotted as a function of time. Initially, the number of decays per time (the slope of the
line) is large, but the rate continuously declines as time goes on.
1,000
Original Number of Radioactive Particles: N0 = 1000
900
# of Radioactive Particles (N)
800
700
600
500
N0/2
400
300
N0/4
200
N0/8
100
0
0
10
T1/2
20
2T1/2
30
40 3T1/2 50
60
70
80
90
100
Time (seconds)
Figure III-10. A typical exponential decay curve, with a half-life (T1/2) of about 14 seconds.
We can’t characterize this curve by its slope, like we do for a linear curve, because it is
constantly changing, but we can describe the curve by its initial value (N0) and the rate at
which it changes. This is an exponential decay curve. An exponential decay curve has a very
interesting characteristic – the fraction of particles decaying is constant for a given length of
time. In the example above, we start out with N0 = 1,000 particles. The time for half of the
particles to decay – about 14 seconds in this case – is called the half-life, and is denoted by
the symbol T1/2. At the end of one half-life, half of the original particles are remaining, so we
have 500 particles left. 14 seconds later, after another half-life, half of the 500 particles are
remaining, so we have 250 particles left. At the end of yet another 14 seconds (t = 42 seconds),
we experience yet another particle decline of 50%, so 125 particles are remaining. This
continues to occur until so few particles are left that the statistical methods upon which
exponential decay is based fail. Looking at the graph, we see that at T1/2, N=N0/2, at 2T1/2,
N=N0/4, at 3T1/2, N=N0/8 and so on. This property of exponential decay allows us to determine
the half-life of a particular decay from any two points on the graph, just like we can determine
the equation for a line based on any two points along the line.
86
The equation for an exponentially decaying curve is of the form:
t
N  1  T1 / 2
 
N0  2 
III-7
where N = current amount remaining after an elapsed time = t
N0 = original amount at time t0
and T½ = half-life.
We can use this equation to solve for the original amount, N0, or the current amount, N, if we
have the elapsed time and the half-life..
We can also determine the half-life of an element by re-writing Eq. III-7. Taking the logarithm
of an exponent gives us just the exponent. If we take the logarithm (“log” on your calculator) of
both sides, we get
 N
log
 N0
t


T1 / 2

1



  log   
 2  



III-8
Now we’ll use an identity of logarithms – an exponent inside a logarithm can be pulled out as a
multiplying factor:
t


T1 / 2
 t
1



log     
 2    T1 / 2


 1
 log 
  2
III-8
so we can re-write equation III-8 as
 N   t  1
  
 log 
log
N
T
 0   1/ 2   2 
III-9
Now we can solve for the half-life or the time. Solving for the elapsed time, we can write the
equation as



 log N  
N 

 0
t  T1 / 2 

 log 1  

 2  

87
III-10
Solving for the half-life we get
T1 / 2


 log 1  

 2 
 t

 log N  
N 

 0 

III-11
Every species of radioactive element has its own unique half-life, which makes its decay rate
extremely predictable. The decay rates of radioactive elements are used to determine the age of
materials. We measure the amount of radioactive decay of a particular element to determine the
species and how much of the decay product is present, and calculate how long it has been
decaying. For example, with every breath we take in radioactive carbon-14, so the ratio of
radioactive carbon-14 to stable carbon-12 in our bodies is the same as the atmospheric ratio.
When we die, we stop accumulating carbon-14 and it decays according to its half-life of about
5,700 years. By measuring the ratio of carbon-14 to carbon-12 in dead animal or plant tissue, we
can determine how long ago it died. This works for objects up to about 60,000 years old.
Beyond that we need to use other radioactive isotopes.
The following table lists the half-lives of some radioactive isotopes commonly used in agedating:
Parent
Isotope
Stable
Daughter
Parent HalfRange of
Life (years) Dating (years)
Carbon-14
Nitrogen-14
5,730
few tens to
70,000
Carbon-bearing remains of animals
and plants
Potassium40
Argon-40
1.3 billion
50,000 to 4.6
billion
Volcanic igneous rocks; micas,
feldspars, hornblende
Rubidium87
Strontium-87 48.8 billion
10 million to
4.6 billion
Metamorphic and igneous rocks;
feldspars, micas, hornblende
Thorium232
Lead-208
14 billion
10 million
Zircon, U and Th-bearing materials
Uranium235
Lead-207
704 million
4.6 billion
Uranium238
Lead-206
4.5 billion
Primary Material Dated
The following examples will give you a chance to see how exponential decay rates are
calculated:
88
Example III-7: 4.00 g of a radioactive element with a half-life of 21 days are left to decay for 45
days. How much of the material is left? We can use the mass as the number of atoms:
N0 = 4.00 g
T1/2 = 21 days
t = 45 days
N=?
 45days 


t
 1  T1 / 2
 1   21days 
N  N0  
 4.00g  
 0.906g
 2
 2
Example III-8: 38 kg of a radioactive material is put in storage for five years. At the end of that
time, only 12 kg remains. What is the half-life of the material?
Less than half of the material remains, so we know the half-life is certainly less than 5 years, but
let’s use Eq. III-9 to get a more accurate value:
N0 = 38 kg
N = 12 kg
t = 5 yrs
T1/2 = ?
T1 / 2




 log 1  
 log 1  


 2  
 2 
 t
  5years
  3.01years




12kg
N
 log

 log

 38kg  
 N  




 0 


After 6 years (two half-lives) only ¼ of the original amount, or 9.5 kg, would remain.
Radioactive isotopes aren’t the only examples of exponential decay and growth. Examples of
other phenomena that obey this relation are populations of organisms (including people),
damped vibrations, liquids leaking out the bottom of a container, photons traveling through semitransparent materials, heating and cooling rates of materials, and the current and voltage build up
and decay in a resistor-capacitor circuit.
F. EFFECTS OF NUCLEAR RADIATION
Nuclear radiation is hazardous to your health. That is why you see caution or warning signs
around areas where nuclear radiation is present, such as the one shown in Fig. III-11.
89
Figure III-11. Warning sign for nuclear radiation.
To protect ourselves from nuclear radiation, we use materials that absorb or stop the radiation.
This is called shielding.
Alpha particles can be stopped by a single sheet of paper or by several inches of air. Alpha
particles are most hazardous when they are ingested, inhaled or absorbed through open wounds
into the body. Alpha radiation is classified as an internal body hazard.
Beta particles can be stopped by a ream or 500 sheets of paper or just a few millimeters of
metal. Beta particles are also an internal body hazard where they can cause severe damage to the
body's organs.
Gamma rays are vastly more penetrating than alpha or beta particles. They are not particles but
electromagnetic radiation. It takes a thick shield of metal to reduce the intensity of gamma rays
and even then some of the gammas will pass through. Shielding for gammas reduces the intensity
to sufficiently low levels to protect the people around the source areas.
Applications
There are extensive applications of radioactive substances in industry. Naturally occurring
sources, such as cosmic rays and radioactive materials in the Earth, produce a steady background
level. Nuclear power plants produce commercial electric power and power a number of our
Naval ships. Nuclear radiation is used in medicine for both the detection and treatment of cancers
and for x-rays.
Units for measuring radiation
There are several units for measuring radiation.
The activity of a radiation source is measured in units of Curies (Ci) and Becquerels (Bq):
1 Curie = 1 Ci = 3.7 x 1010 disintegrations per second
1 Becquerel = 1 Bq = 1 disintegration per second
Radiation exposure is measured in units of Roentgens:
1 Roentgen = 1 R = 3.3 x 10-10 coulombs/cm3 of air
Absorbed radiation doses are measured in units of rads or Grays (Gy):
90
1 rad = 1 x 10-2 J/kg = 0.01 J/kg
1 Gray = 1 Gy = 1 J/kg = 100 rad
The dose equivalent includes biological effects. The Quality Factor (Q), or Relative
Biological Effectiveness (RBE) relates radiation effects relative to x-ray photons with an energy
of 200 keV. 200 keV is 200 kilo-electron-Volts = 200,000 electron-Volts, and one electron-Volt
(= 1 eV) is equal to 1.602 x 10-19 J, so 200 keV is equal to 3.2 x 10-14 J. One MeV is one million
electron-Volts. Electron-Volts are a common unit of energy for atomic and subatomic particles.
Type of Radiation
10 MeV protons
10 MeV neutrons
10 MeV alphas
4 MeV gamma rays
Q or RBE
2
5
15
0.6
The dose equivalent is the product of the absorbed dose and the Quality Factor (or RBE) and is
measured in units of rems (denoted by rems, although “r” usually means rems) and Sieverts (Sv):
1 rem = 1 rad of 200 keV x-rays
1 Sievert = 1 Sv = 100 rem = 1 Gray of 200 keV x-rays.
1 cSv = 0.01 Sv = 1 rem
Effects of Radiation on Humans
Radiation ionizes molecules in the cell. It damages our bodies by
1) interfering with cell functions
2) creating cell poisons
3) interfering with cell reproduction.
The standard statistical estimates for the effects of a certain dose equivalent on the human body,
based on studies of atomic bomb victims are as follows:
Short Term Effects:
Dose Equivalent
> 30 rem
> 300 rem
> 30,000 rem
Effect
swelling, loss of hair; effects generally disappearing within a few weeks
death within one month
death within one hour
Long Term Effects in Terms of Life Shortening Capability, compared to other hazards:
91
Risk
5 rem/year for 20 years
10 lb overweight
Living in a city
Remaining Unmarried
Smoking a pack of cigarettes a day
Life Span Shortened by:
1 year
1.5 years
5 years
5 years
8 years
A dosage of 5 rem/year increases the probability of having a child with a genetic birth defect by
about 4%.
Most Americans receive a about 0.4 rem per year from background radiation, medical x-rays, etc.
For radiation workers, the national standard is that they receive no more than
1) 5 rem/year (~0.1 rem/week, or ~0.002 rem/hr)
2) 0.5 rem/year for those under 18 years of age
3) 0.5 rem/year if you’re pregnant
4) 0.1 rem/year for the general public
Since nuclear radiation is hazardous to your health, handling, storing, shipping and working with
radioactive substances requires highly skilled, trained and meticulous personnel. They need to
know how to control, detect and measure it and how to protect themselves and others from its
hazards. If you are unsure of possible danger or of what to do, report it to your supervisor
immediately. It is better to be super safe than sorry.
92
PROBLEM SET 8: RADIOACTIVE DECAY (77)
1. Which of the following types of nuclear radiation has the greatest speed in a vacuum?
a. alpha
b. beta
c. gamma d. all have the same speed
2. Which would definitely stop a beta particle?
a. 1 sheet of paper
b. 1 ream of paper
c. 1 foot of lead
d. 2 feet of glass
3. Which of the following has the least rest mass?
a. alpha
b. beta
c. gamma
4. On a scale of 1 to 3, with 1 being the highest and 3 the lowest, rank the rest mass, charge and
penetrability of alpha, beta and gamma radiation.
Rest Mass
charge
penetrability
1 ____________
1 __________
1 ___________
2 ____________
2 __________
2 ___________
3 ____________
3 __________
3 ___________
5. The half-life of a radioactive nuclide is
a. half the time needed for a sample to decay entirely.
b. half the time a sample can be kept before it begins to decay.
c. the time needed for half the sample to decay.
d. the time needed for the remainder of a sample to decay after
half of it has already decayed.
6. The half-life of an isotope of radium is 1600 years. How long will it take for 15/16 of a given
sample of radium to decay?
93
7. The half-life of the alpha-emitter 84Po210 is 138 days. This nuclide is used to power a
thermoelectric cell whose initial output is 1 watt. If the output power is directly proportional
to the number of 84Po210 particles, what will the output be after three years?
8. Three days ago, 1.2 x 1019 atoms of a radioactive isotope were present. Due to decay there are
0.57 x 1018 atoms now. What is the half-life of this isotope?
94
OVERVIEW: LABS 77C AND 77D: NUCLEAR RADIATION
All matter is known to contain varying amounts of energy. This energy is stored in the different
particles that comprise the matter. This matter is composed of chemical elements, which consist
of atoms. Atoms in turn are composed of even smaller particles; protons, neutrons, and
electrons.
It is known that all atoms of the same chemical element are not always identical. Atoms of the
same element will always have the same number of protons, but may have a different number of
neutrons. This results in the atoms having different atomic weights. Atoms of the same element
that have different numbers of neutrons are called isotopes.
Any given combination of protons and neutrons is called a nuclide. Nuclide symbols are a sort
of shorthand that is used to describe different nuclides. For example, the three isotopes of
hydrogen are described with nuclide symbols as follows:
Ordinary Hydrogen..
Deuterium................
Tritium.....................
1
1H
2
1H
3
1H
The subscript in a nuclide symbol always gives the atomic number (Z) of the atom. The atomic
number tells you how many protons are in the nucleus of an atom. As mentioned earlier, the
number of protons is always the same in atoms of the same element. You can tell from the
nuclide symbols for the three isotopes of hydrogen that they all have the atomic number 1. This
shows that they all have one proton in their nucleus.
The superscript in a nuclide symbol always gives the nucleon (or mass) number (A) of the
atom. The mass number tells you the combined number of protons and neutrons in the nucleus
of an atom. In atoms of the same element, the number of neutrons in the nucleus of an atom can
vary. Looking at the nuclide symbols for the three isotopes of hydrogen, you can tell that they
have 0, 1, and 2 neutrons in their nuclei.
The isotopes of certain elements are unstable and are in a process of decay. As they decay they
emit radiation. This phenomenon is called radioactivity, and such isotopes are called
radioisotopes. All elements with atomic numbers greater than 82 (and some with smaller atomic
numbers) possess naturally radioactive isotopes. In addition, artificially radioactive isotopes can
be created by bombarding certain stable isotopes with particles. To date, over 2,000
radioisotopes have been discovered.
A common method of detecting alpha and beta particles and gamma rays is to use a GeigerMueller counter (commonly referred to as a Geiger counter). The nuclear scaler you will use in
this lab has a Geiger-Mueller tube mounted inside its cabinet. A diagram of the tube is shown in
the Fig. III-12.
The tube consists of a metal cylinder containing two electrodes and a gas filling. The positive
electrode (anode) is a thin metal wire in the tube. The wall of the tube acts as the negative
electrode (cathode). At the sensor end of the tube there is a thin window of fragile mica, which
allows radiation to penetrate.
95
Mica Window
Positively Charged Wire (cathode)
Figure III-12. The Geiger-Mueller Tube.
When a ray or ionizing particle enters the tube, it causes the gas to ionize. This causes a pulse of
electricity to be sent to the scaler, which triggers a device to indicate that radiation is present. In
the scaler you will use, the device is a counter. In other scalers the device may be a lamp which
will flash. Still other devices use a loudspeaker which may provide the familiar clicking noise
usually associated with Geiger counters.
All Geiger-Mueller tubes do not operate well at the same voltage because of differences in the
construction and gas filling of the tubes. You need to find the operating voltage of each tube by
plotting a Geiger-Mueller plateau.
In these two labs you will first collect data which will enable you to plot a Geiger plateau. You
will use this to determine the operating voltage of the Geiger-Mueller tube. The radiation
emitted by unstable isotopes consists of alpha particles, beta particles, and gamma rays.
LAB 77C – The Inverse Square Law
As nuclear radiation travels away from its source, its strength decreases in agreement with the
inverse square law, i.e. the radiation decreases as the square of the distance from the source. For
example, if you triple your distance from a radioactive source, the radiation you will receive oneninth of the radiation. After you find the operating voltage of the tube, you will use the scaler to
demonstrate how the strength of the radiation decreases as you move the sample away from the
tube.
LAB 77D – Shielding
The mass, charge and energy of different types of radiation determine what is required to block it.
Alpha particles are helium nuclei, with a relatively large mass (and size) and a +2 charge. It’s
easy to stop because of its size and charge. They move at speeds of thousands of miles per
second, but they can be stopped by a thin sheet of paper or a few centimeters of air. Beta
particles are electrons and also move at fast speeds. The electron is much smaller and doesn’t
carry as much charge, but it still has mass (and size) and charge (-1), so it will also tend to be
attracted to other matter. It is possible to stop beta particles with a few hundred sheets of paper.
They have no mass and no charge because they’re photons. They are similar to x-rays, but have a
shorter wavelength and higher energies. They can penetrate thick lead plating and slabs of
concrete. In Lab 77D you will isolate the tube from alpha, beta, and gamma samples using
96
various thickness of different materials. This will demonstrate the penetrating power of the
different samples, and the radiation absorption qualities of the materials.
97
LAB #77C: NUCLEAR RADIATION - INVERSE SQUARE LAW
OBJECTIVES:
SKETCH:
Data Table 1
Trial
#
High
Voltage
Setting
DATE__________
Table 2
1-minute
Reading
Shelf
Background Reading_____
Shelf
Ratio
1
ratio
Count per Minute
BETA () SAMPLE
1
2
2
3
3
4
4
5
5
6
6
GAMMA ()SAMPLE
7
2
8
3
OPERATING
VOLTAGE: ______________
4
5
6
ALPHA () SAMPLE
2
3
4
5
6
98
Graph for Plotting Counts as a Function of Voltage
15000
14000
13000
12000
Counts per Minute (cpm)
11000
10000
9000
8000
7000
6000
5000
4000
3000
2000
1000
0
200
300
400
500
600
700
800
900
1000
1100
1200
Voltage (V)
Analysis
1. Why was it necessary to plot a Geiger-Mueller plateau?
2. You used three different radioactive sources. In the space below, identify each source by its
nuclide symbol and the type of radiation it emits.
Nuclide Symbol
Nuclide Symbol
Nuclide Symbol
Radiation Type_________
Radiation Type_________
Radiation Type_________
3. As the samples were moved further away from the Geiger-Mueller tube, what happened to the
one-minute count? Were the changes in accordance with the inverse square law? Explain
why or why not.
4. Which sample was in the best agreement with the inverse square law? Which was in the worst
agreement?
Use the five-step method to solve the following problem:
5. Using the equation E=mc2, how many joules of energy are contained in a 3 kg mass?
99
LAB #77D: NUCLEAR RADIATION - SHIELDING
OBJECTIVES:
DATE__________
SKETCH:
Data Table 1
Trial #
1
2
3
4
5
6
7
8
9
High Voltage Setting (V)
1-Minute Reading (counts)
Optimum Voltage Setting: ____________ volts
15000
14000
13000
12000
Counts per Minute (cpm)
11000
10000
9000
8000
7000
6000
5000
4000
3000
2000
1000
0
200
300
400
500
600
700
Voltage (V)
100
800
900
1000
1100
1200
LAB 77D
Data Table 2
Absorption Material
Background Reading _____________
Alpha (cpm)
Beta (cpm)
Gamma (cpm)
None
1
2
3
4
5
6
7
8
9
10
Analysis
1. What sort of absorption material was required to stop the alpha particles?
2. What material stopped the beta particles?
3. Did any of the absorption materials stop the gamma rays?
4. Explain why different materials are required to stop different types of nuclear radiation.
Use the five step method to solve the following problem:
5. How long will it take a material with a half-life of 5.75 years to decay from a mass of 8.00
grams to a mass of 2.75 grams?
101
CHAPTER IV: ENERGY CONVERTERS
A. LEARNING OBJECTIVES FOR CHAPTER IV:
1. Define an energy converter and what it is used for.
2. Solve problems in both metric and English unit systems for the energy conversions in the
labs including energy, power and efficiency of the converter.
3. Demonstrate how electrical energy can be converted to light.
4. Demonstrate how light can be converted to electrical energy.
5. Demonstrate how fluid energy can be converted to electricity (wind generator or steam
engine).
6. Demonstrate how mechanical energy can be converted to fluid energy (pump or fan), or
to electrical energy (generator or alternator).
7. Demonstrate how a thermoelectric generator can turn heat energy into electricity or
electricity into thermal energy.
8. Identify workplace applications where the energy converters studied would be used.
B. GENERAL DISCUSSION
Energy Converters are devices that convert energy from one form to another. An
automobile, for example is an energy converter: it converts the chemical energy available in
gasoline and oxygen into the kinetic energy of motion and the electrical energy that powers your
dashboard and radio. The goal of an energy converter is to convert as much of the energy
from one form to another as is practically possible. We say “practically possible” because,
although we want to convert as much energy as possible, we don’t want to pay too much for it.
Cars could get much better mileage if engines were more efficient energy converters, but better
mileage is not at the top of every consumer’s wish list. Furthermore, no energy conversion is
100% efficient - this is one of the laws of our universe. Energy can neither be created nor
destroyed, but it won’t all get converted into a useable form. Energy is always lost to some form
of resistance or misdirection; be it thermal heating through friction, thermal heating through
electrical resistance, or vibrational losses due to the misalignment of a rotating shaft.
The percentage or fraction of energy converted for useful work is the efficiency (). The
fractional efficiency is given as

Eout Pout

Ein
Pin
IV-1
where Eout (“energy out”) is the converted energy and Ein (“energy in”) is the energy source. The
percentage efficiency is just the fractional efficiency multiplied by 100%:

Eout
P
 100%  out  100%
Ein
Pin
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IV-2
Since power is just energy over time (P=E/t),  can also be calculated using the input and output
power (Pin and Pout), as long as both are measured over the same time period. Energy
converters are also power converters.
Often an energy converter will utilize more than one conversion. The ignition of the gasoline in
the cylinder converts chemical energy into thermal energy. The heated gas mixture expands and
is converted into fluid energy, producing high pressures in the piston chamber. The fluid energy
is converted into the translational mechanical energy that moves the piston. That energy is
converted into rotational mechanical energy as it causes the crankshaft/flywheel and, eventually,
the wheels to rotate. The rubber hits the road and converts the rotational mechanical energy into
translational mechanical energy and you move forward. For the electrical system, the rotational
mechanical energy is converted into electrical energy via the generator. Both processes require
multiple conversions of energy from one form to another, and there are inefficiencies, or “losses”
involved with each one. Turbine engines have a similar set of energy conversion processes.
Figure IV-1. Converting the chemical energy in gasoline and oxygen into the kinetic energy of a
batmobile requires multiple energy conversion processes.
The total efficiency of a multiple conversion energy converter is the ratio of the final output
to the initial input energy, as well as the product of the individual fractional efficiencies:
total 
Eout
 1  2  3  4  ...
Ein
IV-3
Example IV-1. Efficiency of the Batmobile
The Batmobile converts 70% of the chemical energy of its jet fuel into fluid pressure. 80% of
the fluid pressure is converted into rotational energy in the turbine. 97% of the rotational energy
is converted into rotational energy in the wheels, and 93% of the rotational energy in the wheels
in converted into translational energy (forward motion).
a) What’s the total efficiency of this process as a fraction?
1 = 0.70
 2 = 0.80
 3 = 0.97
 4 = 0.93
 total = ?
 total = 1·2·3·4 = (0.70)(0.80)(0.97)(0.93) = 0.505
b) What’s the percentage efficiency?
ηtotal (%)  ηtotal  100%  0.505  100%  50.5%
Any energy source you can think of is an energy converter. Here are some examples of energy
converters, classified according to their original energy system:
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Mechanical Energy Converters
Mechanical energy converters involve the energy of mass and motion. A mechanical energy
converter changes mechanical energy into fluid, electrical or thermal energy. A fan converts
mechanical energy into fluid energy. A generator converts mechanical energy into electrical
energy. Automobile brakes convert the car’s translational kinetic energy into waste heat.
Fluid Energy Converters
Fluid energy is most often changed to linear mechanical or rotational mechanical energy. This can
be accomplished via piston cylinders, turbines and windmills. Fluid motors use turbine blades to
convert fluid energy to mechanical energy. Wind turbines, the hydroelectric dam, the waterwheel
and the hydraulic bucket loader are all fluid energy converters. Fluid resistance and mechanical
friction make these devices far from 100% efficient. The efficiency of wind turbines has increased
to the point where they are now commercially feasible. Wind turbine farms can now be found
across the country, and they increase in number every year. The mechanical energy produced by
fluid energy converters is often changed to another form. For example, wind turbines and
hydroelectric dams use fluid energy to get rotational mechanical energy, which is most frequently
used to turn a generator and produce electricity.
Figure IV-2. An offshore wind farm in Denmark
Fluid energy can also be changed to electrical energy through a device called a
magnetohydrodynamic generator ("magneto" means magnetic, “hydro" means fluid, "dynamic"
means motion). A magnetohydrodynamic generator moves an electrically conductive gas through a
magnetic field, which produces a voltage. The gases produce ionized particles consisting of free
negatively charged electrons and positive gas molecule ions, a state of matter known as a plasma.
Generation of electrical energy is accomplished as the gas expands through the duct and cools. After
cooling, the gas is recompressed, heated and returned to the generator. A magnetohydrodynamic
generator can operate with higher efficiencies than conventional electrical power plants. It has no
moving parts and thus is very reliable. Magnetohydrodynamic generators could become an
important source of electrical energy if the technology improves and the price of other energy
sources makes it more economical.
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Electrical Energy Converters
Electrical energy converters convert electrical energy into some other form of energy. Electrical
energy can be converted to rotational mechanical energy by a motor or to thermal energy by thermal
heating devices or simply as waste heat produced by electrical resistance. Electrical energy is also
converted to light (“photonic energy”) for lighting and instrument displays.
Converting electrical energy to mechanical energy or heat energy is probably the most familiar
energy conversion process. Electrical energy converters power the machines in machine shops and
factories and countless other commercial enterprises. In the home, they are found in refrigerators,
electric stoves, washing machines, clothes dryers, irons, vacuum cleaners, mixers, incandescent
lamps, heaters, hair dryers, alarm clocks and computers. The devices that produce heat are
generally the biggest drain on your energy bill.
Thermal Energy Converters
Thermal energy converters change thermal energy into another energy form. Steam plant boilers
generate heat for heating and electricity. They turn thermal energy into fluid energy (steam), which
is used for heating or for powering turbines for propulsion or generating electricity. Internal
combustion engines convert thermal energy to rotational mechanical energy directly. The
thermoelectric generator is another example of a thermal energy converter. Geothermal energy can
be used to create steam or generate a voltage difference via a temperature difference.
Solar Energy Converters
Solar panels, also known as photovoltaic panels, convert light energy into electrical energy.
Continued improvement in the efficiency of solar cells and in their production methods over the
years has lowered costs enough to make solar power competitive in many markets. “Solar power”
is really “light power” – notice that the “solar powered” calculators work quite well under our
fluorescent lights. Currently solar power is most commonly used on small, self contained items or
on large scales in areas that receive large amounts of sunlight but are far from any electrical grids.
Look for solar power (and wind power) to provide an ever-larger proportion of your power needs in
the future.
105
Figure IV-3. The International Space Station relies on photovoltaic cells for power.
Chemical Energy Converters
Any living being, be it a microbe or a Shop Foreman, is a chemical energy converter. It breaks
down food and extracts the chemical energy to do electrical and mechanical work. Electric
batteries are also a source of chemical energy that can be converted into electrical energy.
Nuclear Energy Converters
Our sun’s energy comes from a process known as nuclear fusion. Small nuclei in the sun like
hydrogen, deuterium and helium are smashed together to form larger, heavier nuclei. In the
process matter is converted into energy and released. It takes hundreds of thousands of years for
the light produced in the sun’s core to find its way to the surface, and only a little over eight
minutes to go from the surface to the Earth. The sun is constantly converting its hydrogen into
helium and its helium into carbon and nitrogen and oxygen. It has enough hydrogen to keep
working in its present form for another 4 or 5 billion years, according to theory. Fusion reactors
on the Earth have not had nearly the success of our closest star. We’ve yet to make a fusion
reactor that produces more energy than is required to keep it running. The technological
challenge is to reproduce the pressures and temperatures at the core of the sun while physically
containing the process. The advantage of fusion reaction is that it is seven times more efficient
than our fission nuclear reactors and it produces no radioactive waste products.
All of our working nuclear reactors depend on nuclear fission, which involves very heavy nuclei
like uranium. When certain heavy nuclei split apart, they release energy. The energy is
generally used to heat a fluid, like water (or sodium metal) to operate a steam engine. Deep
space probes use fission to create a temperature difference in a thermoelectric module, which
produces a voltage difference. Nuclear submarines operate on fission reactor power plants.
Their operation is described below.
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HOW NUCLEAR PROPULSION IN A SUBMARINE WORKS
The Primary Coolant Loops and the Reactor Compartment
The energy source of a submarine's nuclear power plant is a uranium pile (the core) in which the
235-nucleon isotope is heavily enriched. Enriching the fissioning component, U-235, provides
more heat in a smaller space. Neutrons smashing into the U-235 nucleus split it apart ("fission
it") into smaller components. The fission process produces other large isotopes, called fission
fragments, which are relatively unstable, as well as neutrons and gamma rays. The largest of the
fission fragments, like Barium and Krypton, collide with the coolant running though the core
area, heating up the water inside to about 550 ºF. That's how the nuclear power is converted to
thermal energy. The fissioning of U-235 also produces neutrons, which smack into other U-235
nuclei, propagating the fission reaction. The pile is contained in a structure made of a zirconium
alloy that is a poor neutron absorber, so it doesn't impede the fission process. It is about the size
of a garbage can, and is encased in a carbon steel pressure vessel with a stainless steel cladding
to limit radiation leakage. The water piping running through the reactor is part of the port
coolant loop and the starboard coolant loop. Heat from the core is transferred to the water in the
coolant pipes. Inserting rods into the core shuts it down. The rods are made of a material that
absorbs neutrons. Boron and graphite work very well, but the sub reactor control rods are made
of something else. When the rods are inserted manually or, as a fail-safe, by gravity, they absorb
too many neutrons for the fission process to continue, and the reactor shuts down. The
pressurizer is located downstream of the reactor core and upstream of the outlet valve. Inside the
reactor, the two loops join, so only one pressurizer is needed for both loops. It raises the water
pressure to around 2000 psig, to keep the 550 ºF water from boiling. The coolant loops then
route the hot coolant into a steam generator (one for each loop). Inside the steam generator is a
heat exchanger. The coolant loops are separated into smaller pipes, which are surrounded by
cooler water. Heat from the primary coolant loops is transferred to the cooler secondary water in
the vessel, generating the steam (at about 550 ºF and 500 psig) that drives the screw and runs the
generators. The thermal energy in the coolant has now been converted to fluid energy in the
form of high pressure steam. The primary coolant cools down to about 500 ºF in the steam
generator, which is a type of heat exchanger. The primary coolant exits the steam generator and
enters the main coolant pumps (centrifugal pumps), which re-pressurize the flow (adding
mechanical energy to the fluid) and force the cooled water back into the reactor, where it is reheated, completing the recirculation process. The sub runs on both coolant loops at once, but if
say, the port coolant loop must be shut down, the starboard loop alone can run the boat, and vice
versa. The two loops add redundancy to the system.
The water in the primary coolant loops is radioactive, which is why its heat is transferred to the
secondary coolant flow by the steam generator, which prevents the radioactive primary coolant
from mixing with the non-radioactive secondary fluid. The reactor, the primary coolant loop, the
pressurizer, the steam generator and the main coolant pumps are all housed in the reactor
compartment, which is surrounded by foot thick steel and lead walls, ceilings and floors, as well
as boreated polyethylene windows to allow visual inspections of the compartment. The walls
aren't completely solid; electrical cables and the secondary coolant pipes snake through them
rather than run directly through them to prevent radiation leakage. In an emergency the plates
that surround the reactor compartment can be unbolted to access the systems. The radioactivity
in the compartment is due to the by-products of nitrogen-16, produced when neutrons are
absorbed by oxygen-16. The half-life of the N-16 is about a minute, so after 30 minutes,
radiation levels are quite low, but trained seamen can get into the compartment before that. In a
107
real emergency, a worker can spend a short amount of time in the reactor compartment before
being relieved; otherwise, they'll wait 30 minutes if they can.
Figure IV-4. The primary plant for a nuclear submarine’s power system.
The Secondary Flow Loops
The primary coolant in each loop's steam generator (heat exchanger) heats each of the two
secondary flow systems (port and starboard), producing dry, saturated steam (550 ºF, 500 psig,
approximately). Leaving the steam generator, the steam exits the reactor compartment and is
directed to the engine room. Steam from each loop is separated into two flows. One flow path
drives the main engine (again, a steam turbine, known as the propulsion turbine). A reduction
gear slows down the rotation rate from the main engine turbine to the screw. This is necessary
so that the main engine can rotate at a faster, more efficient speed. The other flow path drives
the multi-stage turbine generator, which runs the electrical generator, which provides electricity
for the boat. The flow paths exit the turbines at a reduced pressure, having converted some of
their thermal energy into rotational mechanical energy, and then enter the main condenser. Cool
seawater enters through a hole flush with the hull and flows through tubes before exiting back to
the ocean. The main condenser is a heat exchanger, cooling and condensing the exhaust steam to
liquid water at a temperature of about 100 ºF (depending on the seawater temperature, among
other things) and a pressure of about a sixth of an atmosphere. The condensate pump repressurizes the condensed fluid, forcing it to the main feed pump, which forces the cool feed
water back into the steam generator, completing the cycle.
108
Figure IV-5. A generic description of the steam plant aboard a nuclear submarine.
Energy Conversions in the Nuclear Power Plant:
Primary Plant
Reactor: nuclear to thermal
Steam Generator: thermal to fluid (high-pressure steam)
Coolant Pumps: electrical to mechanical to fluid
Pressurizer: electrical to thermal (water is electrically heated to raise its temperature)
Shielding: nuclear to (waste) heat
Secondary Plant
Propulsion turbine and turbine generators: fluid to mechanical
Screw: mechanical to fluid to mechanical (thrust)
Turbine generator: mechanical to electrical
Main condenser: thermal to thermal (no conversion)
Pumps: electrical to mechanical to fluid
Power Plant Transducers
We’ll cover transducers in Chapter V, so it wouldn’t hurt to know how the sub’s nuclear reactor
is monitored. Differential pressure (D/P) cells in the primary plant measure coolant flow rates
and gage pressure at various points. These D/P cells consist of an expandable bellows connected
109
to a linear variable differential transformer (LVDT). Resistive Temperature Detectors (RTD's)
measure the coolant temperatures and the pressurizer temperature. Neutron detectors consist of
ion chambers that measure gamma rays only, and proportional counters (Geiger-Muller tubes)
that measure gamma rays and neutrons. By comparing the two, the neutron radiation levels can
be determined.
The secondary plant contains differential pressure (D/P) cells to measure the steam pressure,
seawater flow rates in the main condenser, and the hot well (condenser) level pressure. Resistive
Temperature Detectors (RTD's) are also located throughout the secondary flow system.
110
PROBLEM SET 9: ENERGY CONVERTERS
1. The efficiency of an energy converter is the output energy ____________ by the input energy.
2. The efficiency of an energy converter can also be calculated by using the ratio of output to input
_________________________.
3. Energy converter efficiency is important because: (circle the correct answer(s))
A. The energy supply is limited by our present technology.
B. The cost of energy increases if converters are inefficient.
4. A mechanical energy converter that is 80% efficient (or has an efficiency of 0.8) has an output
energy of 1250 watts (= 1250 J). What is the mechanical energy input?
5. The blades of a windmill store 5.8 x 104 joules of rotational energy each second in the spinning
windmill shaft. The shaft turns an electric generator. The generator produces 20 kilo-joules of
electrical energy each second. Calculate the efficiency of the windmill generator.
111
6. A ¼ hp water pump is rated at 75% efficient. The motor shaft that turns the pump rotor delivers
137.5 ftlb of mechanical energy to the pump each second. Calculate the power output of the
water pump in horsepower available to do work in moving water.
7. What is the efficiency of an energy converter that produces 400 J of energy in 3 seconds while
using 1000 J of energy in 5 seconds?
112
C. CONVERTING ELECTRICAL ENERGY TO LIGHT ENERGY
Electric lights convert electrical energy into light energy. The most common types are
incandescent lamps, fluorescent lights and light emitting diodes (LED’s). These lamps are
powered by electricity from the AC power line in your house or by batteries (DC).
How do these lamps convert electrical energy to light? Let us first examine the incandescent
bulb. When electricity is passed through a resistance of any kind, heat is generated. If the heat is
great enough, the resistance gets hot and glows. In an incandescent bulb, electricity is passed
through a thin wire called a filament (usually Tungsten). The wire heats up and incandesces
(emits light due to being heated up). The light produced consists of a full spectrum of
wavelengths centered about the yellow wavelengths. The combination of different colors
produces the bright white light you see when switching on an incandescent lamp. The initial
electricity flowing through the wire filament in an incandescent bulb must be converted into heat
and then to light. A lot of heat energy is wasted and a lot of wavelengths are produced that are
too short or too long for us to see (most of the invisible wavelengths are at infrared wavelengths,
which are too long to see). The waste heat and the light produced outside of our visible
wavelength range make incandescent bulbs relatively inefficient energy converters. Incidentally,
any object puts out a broad range of light wavelengths when heated, and the primary wavelength,
or color, depends on the object’s temperature. This kind of radiation is known as blackbody
radiation. Steel, for example, can be heated until it’s red hot. Heat it more and it becomes
white hot – the primary wavelength shortens from the red region to the yellow region. Heat it
still more and the primary wavelength shortens to the blue region – it becomes blue hot. All
heated objects follow this temperature-dependent pattern. When the steel (or the tungsten in a
light bulb) is heated until it’s the same color as the sun, it is at the same temperature (about 6000
Kelvins). Humans emit blackbody radiation too, but at 98.6ºF we’re so cool that we primarily
emit infrared wavelengths, which are too long to be seen by the eye, but are visible through
infrared goggles and cameras.
In a fluorescent lamp, the conversion of electricity to light involves a different process. The long
tube of the fluorescent lamp is filled with a gas (such as neon, mercury or sodium). The inside
of the glass tube is coated with a phosphorescent material. When you switch on the electricity in
such a lamp, a high voltage is generated across the two ends of the tube. The voltage is
generated by two devices: a ballast and a starter. The high voltage causes an electrical discharge
through the gas. Through this process, the atoms in the gas absorb the energy (they are said to
become excited). These atoms quickly get rid of this energy by emitting light in the ultraviolet
range. We cannot see light in the ultraviolet region. However, when such light strikes the
phosphorus coating in the tube, the phosphorus atoms absorb the ultraviolet light. The
phosphorus then re-emits visible light.
The process of atomic absorption and emission is highly efficient. Therefore the overall
conversion of electricity to light is higher in the fluorescent lamp than in the incandescent lamp.
Efficiency for a fluorescent lamp can be 3 to 5 times that of an incandescent bulb.
Electrical Energy is most often described in terms of kilowatt-hours (kW-hr). Last quarter we
showed that 1 kW·hr is equivalent to 3,600,000 J:
113
 J
1 
1000
W
3600
sec

 s 
1 kW  hr 
 3.60 x 10 6 J


 kW   1 hr   W 




Electrical energy is the product of the voltage difference (V), the current (I) and the elapsed time
of current flow (t):
E = VIt
IV-4
The units combine to become joules:
Volts  Amps  sec  V  A  s 
joules coulombs
J C

 sec    s  joules  J
coulomb
sec
C s
Electrical power (P) is energy expended per time. One joule per second is one watt. The equation
for power is
P = V·I
IV-5
Of course you could convert electrical energy or power into English units but, thankfully, everyone
uses the metric system for electrical systems.
Light energy can be measured with a photometer (“photo” is Latin for light). Photometers measure
the amount of light energy striking a photosensitive surface. The light energy is converted into
electrical energy and read off the meter. The intensity of the light is generally determined in
terms of its irradiance (), which is the power (P) per unit area (A):

P
A
IV-6
The units are those of power per unit area: W/m2, mW/cm2, etc. If you know the receiving area of
the photometer (and we’ll give that to you for the lab), you can calculate the irradiance. Given the
irradiance, you can determine the total light power hitting any surface of known area.
P =  ·A
IV-7
Let’s assume we have a point source of light. A light bulb is not a point source, but it looks more
and more like one the further you are from it. The light that leaves a point source at any moment
expands in a spherical shell. The expanding shell of light gets ever larger and more spread out as
you move further and further away. It turns out that the light intensity diminishes as the square of
the distance from the light source. This is known as the inverse-square law. This is not hard to
prove. The surface area of a sphere is
A = 4r2
IV-8
Combining IV-5 and IV-7 we get an expression for the irradiance in terms of the distance from the
source (r):
114

P
P

A 4r 2
IV-9
The irradiance is inversely proportional to the square of the distance. Thus, if you move twice
as far away from a light source, you only receive one-fourth as much light. If you move five times
further away, you receive one twenty-fifth as much light. The total power of the expanding shell of
light remains the same; it’s just spread over a larger surface.
Example IV-2. Irradiance from a light bulb.
A 75 watt light bulb emits light in all directions (a spherical distribution). If the bulb has an
efficiency of 0.80, what is its irradiance at a) 2 meters and b) 4 meters?
r=2m
a) Pbulb = 75 W
 = 0.80
r1 = 2 m
1 = ?
The radiated power Prad, which does not depend on how far you are from the bulb, can be
calculated given the bulb’s input power and its efficiency at producing light:
Prad    Pbulb  0.80  75W  60 W
As the shell of light expands, its total power is 60 W, but that power is spread over an increasingly
larger area with distance. The light leaving the bulb expands in a spherical shell of area A=4r2.
The irradiance () at 2 meters is
1 
P
P
60 W
W


 1.19 2
2
2
A 4r1
m
42m 
b) r2 = 4 m
2 = ?
115
We can either solve this directly or by comparing it to the answer in a). The new distance (4
meters) is twice as far away, so the new irradiance should be one-fourth of that calculated at a
distance of two meters, or about 0.3 W/m2. The direct calculation gives us:
2 
P
P
60 W
W


 0.298 2
2
2
A 4r2
m
44m 
We can get the same answer by taking the ratios of the distances:
2
r 
W  2m 
W
 2   1   1   1.19 2  
  0.298 2
m  4m 
m
 r2 
2
We can reduce inverse-square losses by installing reflectors on light bulbs. A reflector reflects the
light striking it to send more of the emitted light in a desired direction. Reflectors for spherical
bulbs tend to be parabolic or conical in shape:
Figure IV-6. A parabolic reflector concentrates light, enhancing its irradiance in some areas by
redirecting it through reflection.
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PROBLEM SET 10: ELECTRICAL ENERGY/LIGHT ENERGY CONVERTERS
1. A photometer reading shows that an incandescent bulb gives out 2.5mW/cm2 of light power at a
distance of 75 cm from the center of the bulb. How much light power in watts falls on a spherical
surface that surrounds the bulb if the radius of the sphere is 75 cm? Note the surface area of a
sphere of radius R is A = 4πR2.
2. A photometer reading shows that a cylindrical fluorescent bulb gives out 7.2 mW/cm2 of light at a
distance of 100 cm from the center of the bulb. The tube is 1 m long. How many watts of light
power fall on a cylindrical surface, centered about the tube, if the surface has a length of 1 m and a
radius of 1 m? This is different from a spherical source. At these short distances, the area of the
light is an expanded cylinder, with an area A=2r, where  is the length of the tube.
117
LAB 53: CONVERTING ELECTRICAL ENERGY TO LIGHT ENERGY IN LAMPS
OVERVIEW
In this lab you will use an AC wattmeter (power meter) to measure the input (electrical) power to
both types of lamps. You will not need to calculate the electrical power in this experiment,
because your wattmeter will measure power directly.
You will study the effectiveness of using a reflector with an ordinary light bulb. The light will be
placed a given distance above the bench, with and without the reflector. You will use a
photometer to measure the light power striking a surface of known cross section (area). The
photometer measures light power. From this you can calculate the irradiance (), given the area
of the photometer sensor. You can find the light power reaching a reference surface, such as a
standard sheet of paper. You do this by multiplying the photometer reading by the area of the
reference surface.
You will calculate the ratio of the light received by a sheet of paper for both configurations with
the incandescent lamp. You make the comparisons using the light power that falls on the
reference surface at a given distance from the lamp. The ratio is given by the following
equation:
light power received by the referencesurface Pout
Ratio 

IV-10
electricalpower used by the lamp
Pin
The manufacturer of a 15 watt fluorescent lamp states that it will equal the brightness of a 60
watt incandescent bulb. You will test this by visual comparison of the lighting supplied by the
two lamps. You can’t compare the power using the photometer, because it is insensitive to the
light spectra emitted by the fluorescent light. Most photometers are only sensitive to certain
wavelengths of light. The photometer you’ll use in the lab is most sensitive to red light; it estimates
the total power striking it by assuming the red light it measures is part of a full spectrum of
wavelengths. This works great for an incandescent source because it’s a blackbody radiator, but
fluorescent lights emit discrete wavelengths, and the fraction of red light produced varies from bulb
to bulb. Our photometer cannot accurately measure the light from fluorescent sources.
Calculations for this lab:
For each incandescent lamp find the portion of the photometer reading due to lamps:
Pactual = Ptotal - Proom
IV-11
Find the power per unit area (the irradiance, ) for each lamp (the sensor’s active area is 0.5 cm2)
Find total power received
Preceived 
Power
 Area of Paper    A paper
Area
IV-11
Calculate the ratio of light power received by the paper to the input electrical power of the lamp
using equation IV-10. Note that all the dimensions cancel – the ratio is a pure number.
118
LAB 53: CONVERTING ELECTRICAL ENERGY TO LIGHT ENERGY
OBJECTIVES:
SKETCH:
Date______
TABLE 1
Lamp
Input
Power
Plamp
Photometer
Reading
Ptotal
(mW)
Light Received From Lamp
Proom
(mW)
Photometer
Power/area
(mW/cm2)
8½” x 11”
paper
(mW)
Bare
Bulb
Reflector
TABLE 2
Fluorescent Bulb
Labeled Power________________ Measured Power______________
Observations:
119
Ratio:
Pout/Pin
LAB 53: CONVERTING ELECTRICAL ENERGY TO LIGHT ENERGY
ANALYSIS
1. Not all of the input electrical power is converted to visible light. Describe the other forms of
energy which are produced when an incandescent bulb is connected to a power source.
2. What effect did the reflector have on the amount of energy received by the paper?
3. How did the apparent brightness of the fluorescent bulb compare with the much higher-power
incandescent bulb? Which of the losses described in question #1 above will be less for the
fluorescent bulb than the incandescent bulb?
4. A photometer such as the one you used is more sensitive to certain wavelengths than to
others. It is more sensitive to longer wavelengths (red light). Expensive photometers use a
radiometric filter to correct the readings. Why do you think a visual comparison of the
fluorescent and incandescent bulbs was used instead of measuring with the photometer?
120
D. CONVERTING LIGHT TO ELECTRICITY WITH SOLAR PANELS
"Solar cell" is the popular name given to the photovoltaic cell. This device is made up of
semiconductor materials like cadmium sulfide. It converts near-white light to electrical energy.
The light doesn't have to come from the sun; photovoltaic cells produce electricity from any nearwhite light source that is strong enough. Photons of light striking the semiconductor surface
energize electrons, giving them enough energy to cross the semiconductor band gap and initiate
current flow. The electrons only jump across the gap in one direction, so photovoltaic cells produce
direct current (DC). Single solar cells vary in area from 10 cm2 to 100 cm2 and come in an
increasing variety of shapes. A solar panel is an array of solar cells. These cells are wired in series
to provide a higher output voltage.
No other energy source in the last ten years has increased in efficiency and usage as much as solar
cells (although wind power is a close second). The U.S. is currently the largest manufacturer of
photovoltaic cells, but not the largest consumer; this may change as efficiencies rise while
production costs decline. Solar cells are primarily used in other countries, particularly where
hydroelectric and fossil fuel sources of electrical generation are scarce.
Solar panels powered the electric motor that drove the Solar Penguin one man aircraft across the
English Channel. They provide power for telephone circuits in remote locations and for satellites in
orbit (see Fig. IV-3). Many wristwatches and calculators are powered by photovoltaic cells. Solar
panels like the ones used in the accompanying lab are incorporated into the roofs of houses to
satisfy some or all of the homeowners’ power requirements.
Example IV-3: A satellite, like the Mars Global Surveyor, shown below, is powered by silicon
solar cells. If the cells’ efficiency () is 0.08 and the sun delivers an irradiance () of 0.066 W/cm2
to the panel surfaces, find the total area of the solar cells required to produce 12 watts of electrical
power.


 = 0.08
 = 0.066 W/cm2
Pout = 12 W
A=?
121
First solve for the input power:
Pin 
Pout 12 W

 150 W

0.08
The area is related to the input power and the irradiance:
A
Pin


150 W
 2270 cm 2
W
0.066
cm 2
122
PROBLEM SET 11: PHOTOVOLTAIC CELLS
1. A single solar panel is rated at 18 volts, 225 mA (maximum). How many solar panels connected
in series are needed to deliver 1 kW of power at 225 mA?
2. How much power does sunlight deliver per square centimeter to a photovoltaic energy converter
that is 6% efficient and produces a 110 mA current across a 1000 Ω resistor if the sunlight is
collected on an 800 cm2 area? Note that irradiance =  = P/A (W/cm2).
3. A solar cell array receives 750 J of radiant energy in 3 seconds and has an efficiency of 8.7%.
Determine the average output power during this time period.
123
LAB 52: SOLAR PANELS
I.
OVERVIEW
Photovoltaic cells are used to convert light ("photonic" energy) energy into electrical energy
("voltaic" energy). Each cell consists of a semi-conductor circuit that will produce a voltage and
current flow when the electrons on its surface are energized by light. Light comes in chunks of
energy called photons, and their energy is proportional to their frequency. The photons striking
the surface of the photovoltaic cell must be in the right energy range for the type of semiconductor in the circuit. For most applications, the optimum energy range is somewhere in the
visible range, since that is the most abundant form of sunlight received at the Earth's surface.
That's why photovoltaic cells are often called solar cells. The price of this energy converter is
becoming increasingly competitive worldwide, especially in areas far from a centralized
electrical grid. In 1982, a kilowatt-hour of photovoltaic energy would set you back about a
dollar. Twenty years later a kilowatt-hour cost 25 cents, and the trend continues downward. At
the same time, the efficiency of solar cells has steadily increased. Solar cells are connected in
series to form solar panels. Solar panels are generally connected in series to increase the voltage
of the electrical system. Panels connected in parallel would have roughly the same voltage as an
individual panel, but with a more stable current and a longer lifetime.
In this lab, you will calculate the efficiency of a single solar panel, a pair of solar panels in series,
and a pair of solar panels in parallel.
II. OBJECTIVES
1) Measure the conversion efficiency of a photovoltaic panel.
2) Determine the effect of connecting panels in series and in parallel.
3) Measure the sun's irradiance.
III. EQUIPMENT
A solar panel
small motor
2 digital multimeters (DMM)
one photometer for the entire class
circuit board
small light bulb
wire strips
centimeter ruler
124
IV. PROCEDURE
1) Prepare the circuit as shown below. “M” is for motor, “V” is for voltmeter, “A” is for
ammeter. If there is not enough sunlight, we may need to substitute a small light bulb for the
motor.
SOLAR PANEL
+
-
A
V
M
Note that the DMM measuring voltage must be connected in parallel to the load and the DMM
measuring current must be connected in series. Set them both on their highest ranges, then move
the scales down according to the size of the measurement.
Take the panel and the circuit outside. Orient the panel so it is receiving as much sunlight as
possible, though it may be meager on a cloudy, rainy day. Record the voltage and current in the
Table. Check the values with your instructor to ensure your numbers are reasonable.
2) Re-set the DMM's to their highest scales. Partner up with another group and connect two
panels in series, as shown below. Adjust the DMM's to the appropriate scales and record the
values in the Table.
SOLAR PANEL #1
+
SOLAR PANEL #2
+
-
A
V
M
125
-
3) Re-set the DMM's to their highest scales. Connect the two panels in parallel, as shown
below. Adjust the DMM's to the appropriate scales and record the values in the Table.
SOLAR PANEL #1
+
SOLAR PANEL #2
+
-
-
A
V
M
4) Measure the photovoltaic receiving area of the panels as accurately as possible. Note that the
solar cells do not cover the entire surface of the panel. Record the area of your single panel in
units of cm2. Add the areas of the two panels and record this value as well.
126
V. CALCULATIONS
1) Measuring the Input Solar Power
The solar irradiance (that's power per area) will be measured by a photometer. You can make the
measurement yourself or watch somebody else. The photometer must be pointed where the
daylight is strongest, directly at the sun if it is visible. There is a high and low scale setting. The
photometer measures the light intensity in milliwatts (mW). The measured photonic power must
be divided by the area of the photometer's sensor (0.0246 cm2 with the cap on, or 0.50 cm2 with
the cap off – when it’s not very bright outside) to obtain the irradiance (), where

Pmeter
A meter
Record the meter reading and the solar irradiance () in W/cm2.
The power received by the single panel is calculated by multiplying the irradiance by the total
area of the solar cells on your panel:
Pin    Acells
Record the value for your single panel and for the two panel configurations in the Table.
2) Measuring the Output Electrical Power
Electrical power in a DC circuit is the product of the output voltage and current:
Pout = VoutIout
Use this formula to calculate the output electrical power for the single panel, the panels in series
and the panels in parallel. The current must be converted from mA to A for the power to be in
units of watts. Record the values for output power in the table.
3) Calculating the Efficiency of the Solar Cells
The percentage efficiency () of each of the solar panel configurations is given by the formula

Pout
 100%
Pin
Record the percentage efficiency of each of the three panel configurations in the table.
127
LAB 52: SOLAR PANELS
DATE_________
OBJECTIVES
CIRCUIT
TYPE
SKETCH
VOUT
(V)
IOUT
(mA)
POUT
(W)
SINGLE
PANEL
SERIES
PANELS
PARALLEL
PANELS
PHOTOMETER READING: _______________ mW
AREA OF SINGLE PANEL: _______________ cm2
AREA OF BOTH PANELS: _______________ cm2
CALCULATIONS:
128
IN
(W/cm2)
PIN
(W)

(%)
Analysis
1. Which circuit would you expect to have the highest voltage, and why? Which circuit
actually had the highest voltage?
2. Describe the possible sources of error in your measurements.
3. Which circuit had the highest efficiency? Why do you think this circuit was more efficient
than the other two?
4. Five-Step Problem. A solar panel consists of 64 solar cells connected in series, each with an
area of 4 cm2. The solar irradiance is 950 W/m2. The efficiency of the cells is 0.18. The
output voltage is 6 V.
a)
What is the output power, Pout?
b)
What is the output current, Iout?
129
E. CONVERTING WIND ENERGY TO ELECTRICAL ENERGY
Wind turbines have been used for centuries in Europe to grind grain and to provide mechanical
power for pumping water. Wind turbines have seen relatively little use in the U.S., but recent
advances in turbine technology, coupled with rising energy costs, have resulted in the establishment
of wind turbine farms across the country. The Stateline Wind Farm in Southeastern Washington,
for example, generates hundreds of megawatts of power. Currently the best wind turbines are
produced by the oldest users, the Danes.
The energy that can be derived from wind depends on the efficiency of the windmill and also on the
speed of the wind. Wind speed can be measured with good accuracy (±3%) by using a device
called the pitot tube, which we will use in our fluid experiments.
The kinetic energy of any moving mass, like wind, is given by the equation Ek = ½mv2. The wind is
slowed down as it turns the blades and kinetic energy is transferred to the wind turbine. The mass
of air passing through the turbine blades can be computed if the time (t), wind speed (v), air density
( or w) and blade area (A) are known; then the kinetic energy of the air impacting the blades can
be calculated. Not all of the air's kinetic energy can be captured by the system; this built-in
inefficiency is a universal effect known as the second law of thermodynamics.
The fluid power is given by
Pin 
1
1
Av 3  Q V v 2
2
2
If you’re using English units, you may wish to use weight density instead:
Pin 
1 w
1 w
Av 3 
QV v 2
2 g
2 g
Wind turbines are generally used to turn a generator to produce electrical power. The basic
components of a generator are a magnet (either permanent or electromagnet) and a coil of wire
which can rotate in the magnetic field. The mechanical rotation of the coil is caused by the rotating
wind turbine. The output power of the generator depends upon the strength of the magnetic field,
the construction of the rotating coil and the speed at which it turns. The rotation of the conducting
coil in the magnetic field induces, or causes to be formed, a voltage and current in the coil, which is
transferred through conducting brushes. The electrical energy produced can be transmitted over
large distances to where it is needed. In the process of mechanical rotation of the wind turbine and
the generator, some energy is lost so that the output electrical energy is much less than the original
energy of the wind. Example IV-4 shows the (metric) calculations of efficiency for a wind turbinegenerator system.
Example IV-4. Wind turbine-generator system.
A wind turbine has an 8 meter diameter aluminum rotor, with a total surface area of 12 m2. It
operates in a wind speed of 18 m/s. The mass density of air hitting the turbine blades is 1.29 kg/m3.
130
The generator produces 20 kilowatts of electrical power. Find the efficiency of the wind turbinegenerator system.
A = 12 m2
 = 1.29 kg/m3
v = 18 m/s
Pout = 20 kW
=?
The mass of air striking the rotor is given by
m = vAt
where  is the density of air and v is the wind speed in m/s.
The kinetic energy of the air striking the rotor blades is
Ek 
1
1
mv 2  vAt v 2
2
2
The power delivered to the rotor blades is
1

 vAt  v 2
E
2

Pin  k  
t
t
1
 Av 3
2


3
1
kg 
 m
  1.29 3  12 m 2  18 
2
m 
s 

J
 1 kW 
 45,100  45,100 W 
  45.1 kW
s
 1000 W 
The efficiency of the windmill generator () is

Pout
20 kW

 0.443
Pin 45.1 kW
131
This wind turbine-generator system is 44.3% efficient.
The calculation is pretty much the same in English units, except that instead of mass density () we
tend to work with weight density (w), so we make the substitution

w
g
You’ll need to do this in the first homework problem.
132
PROBLEM SET 12: FLUID ENERGY CONVERTERS
1. A set of fan blades is turned by the wind, causing an electrical generator to produce power. Find
the following:
a. The wind speed = 40.0 ft/s and the area of the fan blades is 1.5 ft2. Find the volume flow rate
Qv.
b. The air density is 0.075 lb/ft3. Find the weight of air passing the fan in one second (t = 1 sec.).
c. Calculate the mass from the weight.
d. Find the kinetic energy of the air hitting the fan blades in one second.
e. Convert the units of energy to joules.
f. If the system is 5% efficient and the output voltage = 6 V, find the output current.
133
LAB 51: CONVERTING WIND ENERGY TO ELECTRICAL ENERGY
OVERVIEW
Over the centuries man has converted wind energy into useful work. The resulting applications
included sails and windmills. Wind turbines that generate electricity have become increasingly
common in recent years due to technological advances that have significantly improved their
efficiency.
In this experiment you will use electrical power and a DC motor with a fan blade attached to its
shaft to produce wind. You will apply this wind to a wind-powered generator and measure the
resulting power. Then you will determine the efficiency of the process of converting wind
(fluid) energy to electrical energy.
Calculations for Lab:
Effective Area of Airflow (Afan):
Afan = (ro2 – ri2) = (.052 - .0182)m2 =
_________ m2
The flow velocity is measured by a pitot tube connected to a manometer. The difference
between the dynamic flow pressure and the static pressure can be correlated to the flow velocity
(recall Bernoulli’s equation from last quarter). The density of the fluid used in the manometer
determines the scale and precision of the instrument. Find the average airflow in the ring for
both minimum and maximum airflow.
vavg = ½ [vavg (top) + vavg (bottom)]
Convert the velocities from ft/min to m/s and enter the values in the lab report.
Calculate mass density of air using the air temperature you measured:

 kg
1.293
kg

 3  _____________ 3

m
 1   0.00367 T( C)  m


Find the mass of air striking the generator fan per second:
mass m
(do for maximum and minimum airflows)
  v avgA fan
time t
Calculate the fluid power, which is the fluid kinetic energy per second:
1
2
 mv avg 
E
2

Pfluid  k  
t
t
Calculate the electrical power, Pin (power supply) and Pout (generator):
P = V·I
134
Calculate the percentage efficiency (fluid to electrical) and overall conversion efficiency:

Pout
E
 100%  out  100%
Pin
Ein
135
LAB 51: CONVERTING WIND ENERGY TO ELECTRICAL ENERGY
OBJECTIVES:
SKETCH:
Date_____
DATA TABLE 1
Condition
of Air
Flow
INPUT
Current
(A)
Air Speed
Voltage
(V)
Va (ft/min)
OUTPUT
Current
(mA)
Voltage
(V)
Lamp Condition
Minimum
Top_____
Bottom______
Dim________
Bright______
Maximum
Top_____
Bottom______
Dim________
Bright______
Air Temperature = ______oC
Effective Area of Airflow = ______________m2
DATA TABLE 2
Minimum Air
Flow
Maximum Air
Flow
_________ft/min
________ft/min
_________m/s
_________m/s
Average Air Flow
Mass per second of airflow
m/t
kg/sec
kg/sec
Kinetic Energy per second of Airflow
PF
J/sec
J/sec
Electrical Power In
(PE)IN
J/sec
J/sec
Electrical Power Out
(PE)OUT
J/sec
J/sec
Conversion Efficiency of Fluid to
Electrical Power
1
_______ %
_______%
Overall Conversion Efficiency
2
________%
_______%
136
LAB 51: WIND ENERGY
ANALYSIS
1. There are many energy conversions involved in this experiment. Identify them, beginning
with the electrical energy input to the system and ending with the light output from the lamp
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
2. The efficiency you found in converting the wind energy to electrical energy is quite low.
Identify the sources of resistance causing energy losses throughout the energy conversion
system.
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
137
F. THERMAL ENERGY CONVERTERS - THE STEAM ENGINE
Thermal energy converters change heat to mechanical, fluid or electrical energy. Common thermal
energy converters include thermostats, combustion engines, turbine engines, thermocouples and
thermoelectric generators.
Thermal energy is contained in the vibrational motion of atoms and electrons. Thermal energy is
called “heat”. Heat transfer to a substance is influenced by the amount of heat (or thermal energy)
the substance can hold. The amount of heat energy depends on its mass, specific heat and the
change in temperature:
H = mcΔT
IV-12
We make widespread use of thermal-to-fluid converters in everyday life. Gasoline or diesel engines
power cars and trucks. Thermal energy from the burning of fuel causes the combustion gas to
expand and transfer kinetic energy to the pistons in the cylinders. The kinetic energy is converted to
rotational kinetic energy and electrical energy.
Much of the electricity we use comes from generating plants that utilize steam-powered turbines.
The fuel produces thermal energy that is transferred to the water inside a boiler. As the water turns
to steam it is discharged, under pressure, through a nozzle. The kinetic energy of the steam is
transferred to the turbines as mechanical rotational energy. The rotating turbine turns a generator
and produces electrical energy.
Figure IV-7. A steam generator.
The steam engine, which started the Industrial Revolution, involves several energy conversions
from input to output. First, a fossil fuel, such as coal or oil, is burned. The fuel's chemical energy is
changed to thermal energy. Then the thermal energy is used to boil water and produce steam,
which is a form of fluid energy. Next, the fluid energy in the steam engine is used to drive a piston.
The moving piston changes linear mechanical energy into rotational kinetic energy of a
flywheel. Finally, the rotational kinetic energy of the flywheel is used to drive belts, shafts, gears or
other force transformers to do a certain job.
At each step in the process, some energy is lost. The overall efficiency of the steam engine is
calculated by dividing the final output energy by the input energy and multiplying by 100%.
138
PROBLEM SET 13: THERMAL ENERGY CONVERTERS
1. When electrical power is measured with a wattmeter, the equation for electrical energy is
_______________________________________.
2. When an object is heated, the formula for the thermal energy required is
________________________________________.
3. A small steam boiler is heated by a 1200 watt electrical heating element. How much electrical
energy is produced in one minute? Give the answer in joules and in calories.
4. If the boiler in question #3 contains 2.0 kg of water, what temperature increase (in C0) will occur
in one minute? (Assume that all the electrical energy is converted to heat energy.)
5. If the water is originally at 20 0C, how long will it take to raise the temperature to boiling
(1000C)?
139
6. An electric power plant has a boiler efficiency of 70% (Chemical energy to fluid energy), a
turbine efficiency of 80% (fluid energy to mechanical energy) and a generator efficiency of 60%
(mechanical energy to electrical energy). What is the overall efficiency of the system?
7. How long will it take a heater to produce 20,000 joules of heat energy if the heater operates at
110 V with 12 A of current? Assume that the heater is 90% efficient.
140
G. LAB 55A and 55B: THERMOELECTRIC GENERATOR
A thermoelectric generator or TE module is an energy converter. The TE module has two
semiconductors made of different materials. It either converts thermal energy to electrical energy or
electrical energy to thermal energy; the process is similar to that for thermocouples. When a voltage
difference is applied to the TE module, a temperature difference occurs across the module.
Conversely, when a temperature difference exists across the module, a voltage difference is
produced.
V  T
T  V
When DC current flows through a TE module, the module draws heat from one surface plate and
moves (pumps) the heat to the other plate. Thus, the TE module acts as an electric heat pump.
The direction of heat flow between the plates in contact with the TE module depends on the
direction of electrical current flow through the device.
Figure IV-8 A Thermoelectric (TE) module.
The amount of heat energy moved depends on the temperature difference (T) across the plates and
how much current is flowing through the TE module. TE modules can be used to heat or cool small
surface areas. The surface could be the stage of a microscope, or the body of an integrated circuit
chip. If the current flows through the module in one direction, the surface is heated. If the current
flows through it in the opposite direction, the surface is cooled.
Snowflakes are examined with microscopes to learn about their structures, but snowflakes will melt
at room temperature. A TE module can be used to keep the snowflake from melting while it is
under the microscope. TE modules can be used to keep bacterial cultures alive by heating the
microscope stage. TE modules can be used to maintain clock crystals at constant temperature to
ensure accurate operation of the clock.
Thermoelectric generators tend to have relatively low efficiencies and higher costs. They are ideal
for small scale refrigeration systems.
141
PROBLEM SET 14: THERMOELECTRIC GENERATOR
1. A thermoelectric generator is used to provide power for a weather satellite. It uses enriched
uranium to provide heat energy and a silicon-germanium thermopile to convert the heat energy to
electrical energy. The uranium provides 6 calories per second heat energy. The thermopile
generates 5 watts of power from this heat energy. Find the efficiency of this generator.
2. An energy converter uses heat energy to produce infrared radiation. This radiation produces
thermal energy that is directed onto the 1.8 cm2 sensitive area of a light-sensitive semi
conducting material. The thermal energy is changed to 51.9 mW of electrical power. The
efficiency of the energy converter is 35%. What is the input heat energy each second?
142
LAB 55A & B: THERMOELECTRIC GENERATOR
OVERVIEW
A thermoelectric generator (or module, abbreviated TE module) is an electronic device that can
convert heat to electrical energy. Also, when supplied with electrical energy, this device can
transfer heat energy from place to place. You will observe how a TE module works for each
case.
A TE module consists of two conductor plates connected by an array of pillars. These are
alternately charged positive and negative. It is these pillars that perform the work of the TE
module.
A TE module can be used to generate electricity in the following manner. The two conducting
plates of the TE module are connected to bodies of different temperatures. The plate on the left
is in contact with a hot body. The plate on the right is in contact with a cold body. Using a
voltmeter, you can measure the voltage across the two wires of the TE module.
This electricity can be used to do work or used to monitor a control system. For instance, in
laser power meters, TE modules are used to measure light. When the laser light strikes a sensor,
energy is absorbed. The sensor is in contact with a TE module. When the energy is absorbed,
the temperature of the sensor (and hence one face of the TE module) rises and causes a
temperature difference across the TE module. A voltage is generated by the temperature
difference. This voltage is proportional to the amount of light energy absorbed by the sensor.
When a current is run through the wires, heat is transferred from one body to the other body.
The direction of heat flow depends on the direction of the current flow. TE modules are used in
this way to heat or cool small surfaces. When used this way to cool, the TE module is often
referred to as a solid state refrigerator. When used this way to heat, it is an oven.
In this lab, half of the benches will use a TE module to generate a voltage from a thermoelectric
generator by creating a temperature difference across the module (Lab 55A), and half will
transfer heat from one side to the other by driving a current through the module (Lab 55B).
Calculations for Lab 55A:
Calculate the power for each entry in the table in watts:
P = V·I
Calculate the average power dissipated in TE module, and record it as Pavg
Calculate the total electrical energy transferred:
Eelec = Pavg·t
Graph the power (y or vertical axis) versus the time (x or horizontal axis)
143
Calculations for Lab 55B:
Find the volume for each water reservoir:
D 2h
V
4
Use the initial and final voltage to find the average voltage:
Vavg 
Vi  Vf
2
In a similar manner, find Iavg
Find the average electrical power dissipated in the TE module:
Pavg = Vavg·Iavg
Calculate total electrical energy input to TE:
Eelec = Pavg·t
Find mass of water in the chambers:
m = V
Calculate T = Tf – Ti, in Co
Calculate the heat gained by the heated water:
H = mcT
(cH2O = 1cal/g·Co)
Convert the calories to joules.
Calculate the percentage efficiency:

H
 100%
Eelec
144
LAB 55A
THERMOELECTRIC GENERATOR
OBJECTIVES:
Data Table
Time
t (min)
DATE______________
SKETCH:
Generated Energy
Temperature
Cold Body Tc
(oC)
Hot Body Th
(oC)
Generated
Voltage
V (V)
0 min
0.5 min
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10 min
Pavg = ________________
E(elec) = ______________________
145
Generated
Current
I (μA)
Generated
Power
P (μW)
LAB 55A: THERMOELECTRIC GENERATOR
ANALYSIS
1. Look at Data Table 1. Compare the readings for the 1½ minute mark and the 8½ minute
mark. What is the temperature difference between the two chambers at these two
points?_______
What is the voltage generated at these two points?___________
Can you find a connection between temperature difference and the TE module's ability to
generate electricity?
2. You know that energy can not be destroyed or created. It can, however, be converted from
one form to another. This is the Law of Conservation of Energy. Explain where the energy to
create a voltage across the TE module comes from. Use your data to support your answer.
Can this energy be used to do work? Why or why not?
___________________________________________________________________________
3. What would happen if the temperature difference between the two TE module plates was
increased at the beginning of the experiment? Would the voltage difference be affected?
How?
4. How much voltage do you think you could generate if you put tap water in both chambers
initially?
_______________
5. Assume the volumes of the chambers were increased, with the contact area between the cells
remaining the same. Would it take more or less time for the voltage readings to go down?
______________________________________________________________________________
6. Sketch Power (Y axis) vs. time (X axis):
146
LAB 55B
THERMOELECTRIC REFRIGERATOR/OVEN
OBJECTIVES:
SKETCH:
DATE___________
Dimension of
Volume of
hole in block: h = _______ Diameter D =___________ hole in block: V = ____________
Data Table
Generated Heat
Time
t (min)
Temperature
Cold Body
Tc (oC)
Applied
Voltage
V (V)
Hot Body
Th (oC)
Applied
Current
I (A)
0
0.5 min
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10 min
Vavg=_________________
Iavg=_________________
147
LAB 55B: THERMOELECTRIC REFRIGERATOR/OVEN
ANALYSIS
1. CALCULATIONS: (Show your work below.)
Average power input: Pavg=______________________________________
Total electrical energy input: Eelec = __________________________________________
mass of water: m =________________________________________________________
Temperature increase of water: ΔT = _________________________________________
Heat gain: H = _____________________________ = _________cal = _________ J
Heating Efficiency: η = ________________________________________
What happened to the rest of the electrical energy?
2. Explain how the TE module can be used as a refrigerator.
3. Suppose the volumes of the two chambers were increased. Do you think the temperatures
would change faster or slower than they did in your experiment? (Assume the amount of
electrical energy available was the same).
4. Which changed more in the experiment: the temperature of the cold chamber, or the
temperature of the hot chamber?___________ Explain what you think this means.
148
PROBLEM SET 15: REVIEW
1. (Fluid to Electrical) A wind turbine with a 30 foot diameter aluminum rotor has a total blade area
of 256 ft2. It operates in a wind speed of 60 ft/s. The windmill generates 20 kilowatts of electrical
power. Find the efficiency of power transfer. Use 0.075 lb/ft3 for the weight density of air (w).
2. (Electrical to Mechanical) A motor is rated with an input voltage of 110 V and a current of 2.5 A.
If the motor efficiency is 0.85, find the output power in horsepower.
3. (Electrical to Thermal) A 240 watt heating coil is used to heat one liter of water in a container. If
90% of the heat is absorbed by the water, find the time needed to raise the water temperature from
25 0C to 90 0C.
149
4. (Thermal to Electrical) A thermal electric generator has two cells each containing 25 grams of
water. The temperatures of the warm and cold cells start at 920C and 200C, respectively. The hot
water has cooled to 750C after five minutes. a) Find the amount of energy, in joules, that the hot
water gave off. b) What is the maximum electrical power that could be produced?
5. (Multiple Conversion) A hydraulic lift system is powered by burning coal. The following energy
converters - at the given efficiencies - are used in the delivery of the necessary energy:
Boiler: 75% Turbine: 82% Generator: 62%
Electric motor: 79%
Fluid pump: 93% Hydraulic cylinder: 96%
Determine the overall efficiency of the system and the weight of coal that must be burned each
hour to provide a power output of 20 horsepower. The energy content of the coal is 13,500 BTU
per pound.
6. (Multiple Conversions) A 100 kg miniature motorized vehicle converts 11% of the energy
created by combusting its fuel into vehicle kinetic energy. What speed can the vehicle obtain in 20
seconds, starting from rest, if the burning of motor fuel liberates 1000 joules of energy each second?
150
CHAPTER V: TRANSDUCERS
A. LEARNING OBJECTIVES FOR CHAPTER V
1. Identify a transducer as a device that senses mechanical, fluid, electrical or thermal
information.
2. Describe the action of a transducer in general terms.
3. Distinguish between an energy converter and a transducer.
4. Identify transducers that change mechanical signals into electrical signals (strain gauges,
accelerometers and microphones); fluid signals into mechanical or electrical signals
(turbine flow meters, Bourdon tubes and rotameters); electrical signals into mechanical or
thermal information (ammeters and voltmeters) and thermal signals into mechanical, fluid
or electrical information (thermistors, thermocouples, bimetallic strips, solid-state
temperature sensors or platinum resistance thermometer).
5. Explain the piezoelectric effect and how it is used.
6. Explain how a Bourdon tube pressure gauge works.
7. Explain how the moving-coil transducer changes electrical input into a mechanical output
signal.
8. Explain why an ammeter has a low-resistance shunt connected in parallel with the
transducer circuit and why a voltmeter has a high resistance resistor in series with the
transducer circuit.
9. Describe what an electrostrictive transducer does and what a photoconductive transducer
does.
10. Describe how the following operate: bimetallic strip, thermometer, thermocouple,
resistive temperature device and thermistor.
11. Solve quantitative transducer problems.
12. Identify workplace applications of transducers.
B. INTRODUCTION
Transducers are devices that convert energy from one form to another in order to provide
information about an energy system or process. Transducers are measuring devices whose
output is often used to provide indication and control information to human operators or
automated systems. They differ from energy converters because they try not to maximize the
amount of energy converted, but rather to minimize the amount of energy converted.
Transducers are used to measure a characteristic of an energy system, so they want to disturb the
system they’re measuring as little as possible. As an example, engine temperature is the input to
a transducer that converts the information to an electrical signal to activate an alarm if safe
engine temperatures are exceeded. A transducer picks up or senses a temperature, a force, a
pressure, a voltage or a flow rate. This input signal is changed into another signal that affects the
reading of a measuring device located at another location.
The output of a transducer will vary as input varies and thus can drive a gauge or a meter. In a
gauge the output is used to drive a mechanical system such as gears and lever arms to produce a
reading. In a meter the output of the transducer is an electrical signal used to produce a reading
on the meter. When talking about gauges or meters you need to know the difference between
analog and digital readings. Both give the same information. In an analog gauge or meter the
pointer moves along the scale in proportion to the strength of the signal. The stronger the input
signal to the transducer the stronger the output of the transducer is to the gauge or meter input.
In a digital meter the output signal of the transducer produces a numerical reading on the meter.
151
A digital reading can be read directly, while an analog reading requires interpolation in order to
get the reading.
Another characteristic of a transducer is that it can transmit as well as transform a signal. If they
derive their power from the input system they need no external energy source for their operation.
Examples of transducers:
Mechanical: strain gauges, microphones, accelerometers, phonograph needles, float gauges,
position gauges and barometers. They change mechanical inputs to other forms of energy.
Fluid: turbine flow meters, underwater hydrophones, rotameters and Bourdon tubes. They
change fluid inputs to other forms of energy.
Electrical: Ammeters and voltmeters. They change electrical inputs to other forms of energy.
Thermal: thermocouples, thermistors, bimetallic strips, thermometers and resistive temperature
devices. They change thermal inputs to other forms of energy.
Optical: photocells and charge coupled devices (CCD’s). They change light energy into
electrical or other forms of energy.
Transducers are important in our modern society. They take an input signal and make it
available to a controlling mechanism, which may be an operator or a computer. Most machines
made today include transducers. They allow proper operation of the equipment and improve
productivity. Sometimes the output is simple, such as a gas gauge, which lets us know when to
fill the gas tank. Other times the output is vital to safe operation of the equipment, such as a
nuclear power plant gauge monitoring fluid cooling flow. Without proper flow the reactor can
overheat, with potentially catastrophic results.
The output of a transducer is always related to its input. As temperature increases, the voltage
output of a thermocouple increases, as fluid flow increases, the counting rate of a turbine flow
meter increases. A transducer gives a quantitative indication of its input.
C. FORCE TRANSDUCERS
Introduction
In the study of the strength of materials, we find that the shape of materials subject to forces can
change. The force applied per area causes stress while the deformation or change in the shape of
the material indicates strain. A strain gauge measures the stress on an object by measuring the
change in resistance as the gauge stretches or compresses. The induced strain can be used to
determine the amount of stress applied. Wire, foil, ceramics and semiconductors are a few
examples of materials than can be used in strain gauges. Their shapes, sizes and sensitivities
vary; each can be used for certain applications.
The resistance of a wire depends upon its material, its cross sectional area and its length. The
formula (from Physics 171) is:
R = 
A
where R = resistance
 = resistivity of the material
 = length of the conductor
A = cross sectional area of wire = d2/4
V-1
When the wire is stretched it becomes thinner – its area decreases and its length increases –
which, from equation V-1, will cause its resistance to increase. When forces are applied to an
152
object that is not free to move in response, its shape can change. The change in shape may be
large or small and either temporary or permanent for the material. Figure V-1 shows the
different ways forces may be applied to an object and the resulting changes in shape. The types
of responses in shape to stresses are compression, stretching (or elongation), shear, torsion and
bending.
torsion
compression
stretching
shear
Bending
Figure V-1. Types of forces and how objects change shape.
Arrows indicate directions of applied forces.
Modulus of Elasticity
Elasticity is a property of materials. Highly elastic materials will change their shape under stress
but return to their original shape when the stress is removed. Inelastic materials will similarly
change shape but will be permanently deformed when the stress is removed. For example,
molding clay is a relatively inelastic material, while rubber is relatively elastic.
Young’s Modulus of Elasticity (Y), developed by Thomas Young in the 19th century, describes
the response of a material to an applied stress. It relates the stress applied to a material to the
strain it produces:
F
stress A
F
Y


strain  A  

V-2
where the stress is the applied force F divided by the contact area A, and strain is the change
in length  divided by the unstressed length . Typical units for Young's modulus are psi
(lb/in2) in the English system and N/cm2 for the SI system. Fig. V-2 is a graph of a typical
response (strain) of a material undergoing stress.
153
Figure V-2. Typical stress versus strain curve.
When stress is first applied the change in shape or the strain increases correspondingly. Below
the elastic limit the material returns to its original length if the stress is removed. If the stress
continues above the elastic limit the material will be permanently deformed. If the stress
continues to increase, the breaking point will be reached and the material will fracture. The
Young's modulus for various materials can be found in the reference tables.
Hooke's Law, which we studied earlier, is F = kd where F is the force applied, k is the Hooke's
Law constant for the material and d is the change in length. If we rearrange Young's Modulus
equation to solve for F we will get:
F = YA

V-3
If we set k=YA/, this is just Hooke's Law, therefore Young’s modulus (Y), stress, (F/A), and
strain, (/), are all related in a formula that controls the stretch of an object. All objects behave
like springs to some extent.
The Strain Gauge
A strain gauge is essentially a variable resistor. An applied stress changes the resistance of the
gauge by changing its length and cross sectional diameter. From the equation for resistance,
R=/A (eq. V-1), we see that if the object is stretched, causing its length to increase and its
diameter to decrease, its resistance will increase. Conversely, if the gauge is subject to
compression the length will decrease and the diameter will increase, causing the resistance to
decrease. Figure V-3 shows a set of metal foil strain gauges.
154
Figure V-3. Typical metal foil strain gauges
Strain gauges are used for force measurements. A load cell is a strain gauge that is rugged,
sensitive and accurate. A commercial load cell rated to measure loads up to 20,000 lb can
measure the weight of the average adult to within 10 lb or less. Loads cells are commonly used
to weigh cargo, airplanes and trucks.
Example V-1. Change in resistance when a wire is stretched.
A 1.0 mm diameter gold wire is 4.0 meters long. What is its change in resistance when it is
stretched 4.0 cm and the diameter shrinks to 0.9 mm? Looking in the back of the book, we find
the resistivity of gold to be 2.44 ·cm.
 = 2. 44 ·cm = 2.44 x 10-6 ·cm
1 = 4.0 m = 400 cm
 = 4.0 cm
d1 = 1.0 mm = 0.10 cm
d2 = 0.9 mm = 0.09 cm
R = ?
The original resistance is R1:
R1 


 1
 1
2.44x10 6   cm 400 cm 


 0.124 
2
A  d1 2 
0.10 cm 


 4 
4


The new resistance is R2, and 2 = 1 +  = 400 cm + 4.0 cm = 404 cm:
R2 


 2
 2
2.44x10 6   cm 404 cm 


 0.155 
2
A  d 2 2 
0.09 cm 


 4 
4


The change in resistance R is given by:
R = R2 - R1 = 0.155  – 0.124  = 0.031 
155
PROBLEM SET 16: STRAIN GAUGES:
1. An aluminum wire is run between two buildings. A) If the wire is 104 cm long and 0.01 cm in
diameter, how much electrical resistance can be expected? B) If I pulled on the wire with a force
of 600 newtons, how much would it stretch?
2. In problem 1 above, after 6 months the wire is found to have increased its length by 5% and its
cross-sectional area has decreased by 55%. Will the resistance change? Will it increase or
decrease? Explain your answer.
3. The resistance of a wire is 10 ohms when the wire has a length of 2 m and a cross-sectional
area of 0.003 m2. What is the resistance of the wire when it is stretched to a length of 3 m and
has a new cross-sectional area of 0.002 m2? Assume the electrical resistivity is unaffected by
the strain.
4. A 300 lb pumpkin hangs from a 0.25 inch diameter steel wire that is 60 ft. long. How much
does the pumpkin stretch the wire?
156
LAB 57: FORCE TRANSDUCERS (STRAIN GAUGES)
OVERVIEW
A force transducer is used for measuring forces. There are various types of force transducers,
which operate on different principles. In this experiment you are going to use an electronic load
cell and a strain gauge as a force transducer. The load cell consists of a 6" long, 2.25" OD (outer
diameter) PVC tube with an attached printed circuit board. The strain gauge is mounted to the
side of the load cell.
Using a pressure stage and hydraulic jack, you will subject the load cell to a stress in the form of
compression. Stress is equal to the force (F) applied to an object divided by the area (A) to
which the force is applied.
Stress = F/A
Since PVC is somewhat elastic, the load cell will change slightly in length under stress. Because
you are applying a compressive stress to the load cell, it will shorten slightly. This shortening is
a strain on the material of the load cell. Strain is equal to the change in load cell length ()
divided by the unstressed length of the load cell ().
Strain = /
Given the stress and strain, you can solve for Young's modulus (Y). Young's modulus is defined
as stress divided by strain:
Y = Stress/Strain = F/A
In this lab, a force will be applied to a load cell. In this particular lab, the force (F) will be
calculated by multiplying the jack pressure by the area of the jack piston. The area (A) in the
equation above is the cross sectional area of the circular ring of the cell. The length () is the
height of the cell. The strain will not be measured directly by the very small change in height,
but instead it will be indicated indirectly by the change in electrical resistance of a wire.
The compression or stretching of the load cell affects the resistance of the strain gauge that is
mounted to it. When you change the length of the load cell, you also change the resistance of the
wire in the strain gauge. When a wire is stretched, its resistance will increase, because its length
increases and its cross-sectional area decreases. Under compression the wire's length will
decrease and the cross-sectional area will increase, causing the resistance to decrease.
You will subject your load cell to varying compression stresses. Using a digital multimeter
(DMM), you will observe the resulting changes in current flow through the wire of the strain
gauge. Then you will graph a calibration curve for your load cell using the compression stresses
and meter reading.
157
LAB 57
FORCE TRANSDUCERS (STRAIN GAUGES)
Date_________
OBJECTIVES:
SKETCH:
Reading #
Pressure Gauge
Reading
P (psi)
0
0
1
2
3
4
5
1000
Current Through Strain Gauge
I (A)
Graph of Force vs. Current
I (A)
F (lb)
158
Force Applied to
Load Cell
(Calculated)
F (lb)
LAB 57: FORCE TRANSDUCERS
Table 2
Piston Diameter
mm
inch
Area
(in2)
Intermediate Point Readings
I (a)
P (psi)
Force (lb)
From Graph
Calculated
QUESTIONS:
1. If you could stretch your load cell, describe the effect on the resistance of the strain gauge,
and explain.
_______________________________________________________________________
______________________________________________________________________________
2. Lowest Load Value = Fmin = _____________lbs
3. Compute the percentage of difference between the calculated force and the force estimated
from the graph.
______________________________________________________________________________
4. Describe how your load cell and apparatus could be used to measure an unknown
compressive force.
______________________________________________________________________________
______________________________________________________________________________
5. The load cell is made from material that expands as its temperature increases. Describe the
effect on the resistance of the strain gauge when the temperature increases.
______________________________________________________________________________
______________________________________________________________________________
6. Why is it necessary to "zero" the gauge every time it is used?
_____________________________________________________________________________
7. A 27 inch long wire is made of steel which has a Young's modulus value of 3x 107 psi. The
wire has a diameter of 0.25 inch. If a weight of 500 lb is suspended from the wire, find its
change in length.
159
D. VIBRATION TRANSDUCERS (LAB 58)
Microphones
A microphone is a common device for converting a mechanical vibration (sound) into an
electrical signal. There are several different types of microphones, each of which employ
different principles in their operation. One common microphone is the carbon microphone,
which contains a diaphragm that moves or vibrates in response to sound waves hitting it. Under
the diaphragm is a small cup filled with carbon grains. A low voltage-driven current passes
through the carbon grains. As the diaphragm moves in response to the incident sound waves it
will compress or relax the pressure on the carbon grains. The resistance of the carbon grains
changes as they are compressed or released from pressure by the diaphragm. This changing
resistance changes the amount of current passing through the carbon. Highly compacted carbon
grains have less resistance and thus a higher current passing through them. The diaphragm and
the carbon grains form a mechanical transducer, changing acoustic waves into varying current
representing the input signal.
A capacitance microphone is shown below in Fig. V-4. Sound waves cause the flexible front
capacitor plate to oscillate. The front and back plates form a capacitor, and the oscillating
distance between the two plates produces an electrical signal in the form of a varying voltage.
Figure V-4. Schematic of a capacitance microphone
A dynamic microphone is shown below in Fig. V-5. Sound waves vibrate the diaphragm,
causing it to move back and forth. The diaphragm is connected to a coil, which surrounds a
permanent magnet. As the coil moves back and forth, the permanent magnet induces a current
through it, converting the audio signal into an electrical signal. This can be converted back into
an audio signal by running the dynamic microphone in reverse. When run in reverse, it’s a
standard speaker.
Figure V-5. A dynamic microphone. Operated in reverse, it’s a speaker.
160
The Piezoelectric Effect
Certain types of crystals exhibit a phenomenon called the "piezoelectric effect" when mechanical
stress is applied to them. These crystals develop an electrical charge on their surface when
pressure is applied to them. The atomic structure of a quartz crystal is shown below in Fig. V-6.
The positively charged, larger atoms are silicon, while the smaller, negatively charged atoms are
oxygen. In their unstressed state, on the left, there is no net charge at the surface. On the left
wall, for instance, the positively charged silicon atom closest to the wall is balanced by the two
oxygen atoms, which provide twice as much charge, but are far enough away from the wall to
just cancel the silicon atom’s charge. Along each wall of the crystal section shown, the charge is
neutral. If we squeeze the crystal however, we get the configuration shown on the right side.
Now there is a net negative charge on the left side, and a net positive charge on the right side.
By applying a pressure difference to the crystal, we have produced a voltage difference! This
effect is known as the “piezoelectric” effect, from the Greek “piezo”, meaning “to press”, and the
Greek “electric”, meaning, well, “electric”.
-
-
+
+
-
-
+
+
-
+
-
+
+
Figure V-6. The piezoelectric effect in a quartz crystal. Negatively charged atoms are oxygen,
while the positively charged, larger atoms are silicon. The left side is the uncompressed state
and the right side is the compressed state.
Application of pressure results in strain on the crystal and causes electrical charges of opposite
polarity to form on the two surfaces for a short time. When the pressure is released the charge
disappears quickly; the atoms will reorient themselves to reduce or eliminate the V. Thus, a
piezoelectric crystal is a transducer that changes mechanical energy into electrical energy. The
amount of charge or voltage created is directly proportional to the amount of strain the crystal is
undergoing. Conversely, by applying a voltage across the crystal, we can induce a strain and
change its shape. We’ll get into this more when we discuss sonar.
These crystals are used in dynamic situations, that is, where conditions are changing. Examples
include certain microphones, accelerometers and underwater hydrophone detectors. The
microphone has a crystal that responds to the changing pressure of acoustic signals (sound),
producing a voltage. Crystal microphones use Rochelle salt (sodium potassium tartrate) or
quartz to provide higher quality sound reproduction than the carbon microphone mentioned
above.
There is a table of piezoelectric properties in the reference tables.
161
Accelerometers
Accelerometers are devices that detect or sense acceleration or changes in motion. An
accelerometer uses a mass to create pressure on the piezoelectric crystal, as shown in Fig. V-7.
Changing the rate of motion (acceleration/deceleration) changes the force that the mass applies
to the crystal. The stress on the crystal produces a voltage that is proportional to the amount of
acceleration.
Spring-Loaded
Calibrated Mass
Piezoelectric crystal
“doughnut” disk
V
crystal mounting
Figure V-7. A piezoelectric accelerometer configuration.
One formula used with piezoelectric crystals is:
V = SP
V-4
where V is the potential difference, P is the applied pressure and S is sensitivity of the crystal
(in mV/psi, for example). The sensitivity is given by S = kh, where k is the piezoelectric
constant (voltage produced per meter for each N/m2 of pressure applied, for example) and h is
the thickness of the crystal. Values of k are listed in the reference tables.
Another formula used with piezoelectric crystals is:
V = Sga
V-5
where a is the acceleration in “g's” and Sg is the g-sensitivity (in units of mV/g, for example).
Equation V-5 is really a restatement of V-4. Pressure is force per unit area, and force is mass
times acceleration, so
162
P 
F ma
Sm khm

, thus S g 

A A
A
A
V-6
You have probably felt vibrations from a machine (like your car or blender) while it was
operating. Certain levels of vibration are considered normal. Sometimes, however, old or worn
machines vibrate excessively (due, for example, to a misaligned shaft), which can cause
excessive wear or break down. The ability to detect and measure the vibrations of an operating
piece of equipment can help the operator detect problems before a costly breakdown occurs.
Accelerometers are vibration transducers that sense motion and produce an electrical signal that
allows these motions to be analyzed. They detect changes in motion in a single direction but, by
mounting more than one at a time, multiple directions can be measured at once.
In the introduction to this unit we learned that accelerometers depend upon the piezoelectric
crystal. The accelerating mass presses on the crystal, causing an electric charge to develop on the
face of the crystal. The greater the acceleration, the greater the force (F=ma) on the crystal,
which causes more charge and a larger output signal from the crystal. Note: This is true only
when the direction of the vibrating body is perpendicular to the surface common to the
mass and the crystal. Motion parallel to the common surface produces little or no charge
on the crystal and hence no output signal from the transducer.
This directional sensitivity of an accelerometer allows us to analyze the vibrations of machines in
different directions. One transducer can be mounted for up-down vibrations while another can
be oriented for right-left vibrations
163
PROBLEM SET 17: PIEZOELECTRIC CRYSTALS
1. A device used to sense the conditions of a system is called a ________________________.
2. Certain crystals develop a charge on their surface when a pressure is applied to them. This is
referred to as the ___________________________________.
3. A quartz crystal transducer is used to measure impact forces of automobiles in collisions. The
transducer has an operating range of 0 - 80,000 psi, with an output voltage range of 0 - 6 volts.
The sensitivity of the crystal is 0.075 mV/psi. When a force of 12,000 lb acts on a 1 square
inch area, what output voltage does the transducer produce?
4. How much pressure must be applied to a piezoelectric crystal of lithium sulfate 5mm thick to
produce a potential difference of 200 V?
5. In an accelerometer, a piezoelectric crystal generates an electric output when
_______________________________________________________________.
6. An accelerometer with a sensitivity Sg of 14 mV/g (g is the acceleration due to gravity)
produces a voltage of 0.55 V when in contact with an engine. What is the acceleration?
164
7. An accelerometer with a sensitivity Sg of 12 mV/g is used to measure an acceleration of
2.6g’s. What voltage is produced?
8. What is the sensitivity Sg of an accelerometer that produces a voltage of 10 mV when
measuring an acceleration of 0.15g?
9. If an accelerometer has the same sensitivity as in the previous problem and is used to measure
an acceleration of 25 m/s2, what is the voltage produced?
165
LAB 58: VIBRATION TRANSDUCERS
OVERVIEW
When riding in an automobile, you feel comfortable as long as the vibrations are very low.
When the vibrations become excessive, it is time to consult a mechanic. The excessive
vibrations usually indicate that something in the automobile is about to break down.
In a machine shop, operators feel - as well as hear - vibrations from the machines. From these
vibrations, they can tell when a drill bit or machine tool is dull. From vibrations the experienced
operator can determine what force to apply to a tool.
As parts wear in a machine, excessive vibrations can result. Not only can worn parts cause the
machine to lose accuracy, they may also cause the machine to break down.
Although you may be able to hear and feel vibrations, there are limitations that your hearing and
sense of touch have for interpreting the data. With the aid of vibration transducers, you will be
able to further analyze these vibrations. The vibration transducers produce an electrical signal
that is proportional to the amplitude and frequency of the mechanical vibrations. The vibration
transducers, or accelerometers, that you will use in this experiment depend on the piezoelectric
effect for their operation.
In this experiment you will use two accelerometers mounted perpendicular to each other. The
accelerometer mount on the vibration table permits you to vary the angle of the accelerometers
with respect to the vibrations of the table.
During the experiment you will hold a small object where the vibrating table will hit it
repeatedly. The object that is recommended is a 9/64" Allen wrench. Compared to the mass of
the table, it is relatively small. It should cause the table to vibrate at a frequency other than the
vibration caused by the motor. This can be seen in the two plots below (Fig. V-8), where we see
an additional vibration on top of the main vibration of the motor. On the left side the wrench is
above the platform, and hitting it on the upstroke. In the plot on the right, the wrench is below
the platform and hitting it on the downstroke. Your oscilloscope readouts may not look this
clean, but at least now you know what to look for. After you have performed the experiment
using the 9/64" Allen wrench, hold other objects such as screwdrivers or pencils against the
vibrating table and observe the result.
1.5
1.5
1
1
0.5
voltage
voltage
0.5
0
0
2
4
6
8
10
12
0
0
-0.5
-0.5
-1
-1
2
4
6
8
10
12
-1.5
-1.5
time
time
Figure V-8. Piggy back vibrations of an Allen wrench on top of the main platform oscillations.
How would the vibration transducer you are using interpret these vibrations? The other name for
the vibration transducer is accelerometer. This indicates that the output of the accelerometer is
proportional to acceleration rather than position.
166
LAB 58
VIBRATION TRANSDUCERS
DATE____________
OBJECTIVES:
VIBRATION PERIOD (T):
 TIME 
T  # DIV 
  ______  _______  ________
 DIV 
TRIAL #2 Noise with Allen wrench
held against bottom of table
TRIAL #1
TRIAL #3 Noise with Allen
wrench against top of table
TRIAL #4 Accelerometers in diagonal position #1
167
TRIAL #5 Accelerometers in position #2
LAB 58: VIBRATION TRANSDUCERS
ANALYSIS
Horizontal division of one cycle:_______________________
CALCULATED PERIOD AND FREQUENCY:
A. From the known oscilloscope setting, compute the time of one cycle. This is the period, T.
B. Find the rotating velocity of the motor shaft, in Hz. f = 1/T
C. Convert revolutions per second to rev. per minute.
Does this frequency agree with the strobe reading?____________
QUESTIONS
1. When you held the Allen wrench against the bottom of the table, did the first instance of noise
occur while the sine wave was going positive or going
negative?____________________________
2. Assume the vibration of a machine is in one direction only. How should the face of the
transducer be mounted to give the best display on the oscilloscope?
3. Assume the vibration of a machine is in one direction only. You mount one transducer on
the machine to give you maximum output. You mount the second transducer with its contact
face perpendicular to the face of the first transducer. What signal does the second transducer
produce on the oscilloscope?
______________________________________________________________________________
4. The amplitudes and frequencies of the waveforms in Trials #4 and #5 should have been the
same. Describe the characteristics that did change between Trials 4 & 5.
______________________________________________________________________________
5. A transducer is mounted on a machine with its contact face parallel to the floor. A second
transducer is mounted with its contact face perpendicular to the floor. When the machine
vibrates, it produces identical signals from both transducers. The signals are in-phase. In the
figure, draw the direction in which
the machine vibrates and indicate
its angle in respect to the floor.
Accelerometer A
Accelerometer B
Machine
Floor
168
E. FLUID FLOW TRANSDUCERS
Fluid transducers are information collectors in fluid systems. They detect the condition of a gas
or liquid in a fluid system. Just like mechanical transducers, fluid transducers change the input
signal to an output signal. Usually the output signal is in the form of mechanical or electrical
energy. Fluid transducers most often measure pressure or flow rates in fluid systems. Often they
are used to measure liquid flow rates in a pipeline. They are also used to monitor operating
conditions inside pneumatic and hydraulic systems. Turbine flow meters, Bourdon tubes,
barometers and anemometers are typical fluid transducers.
Turbine Flow Meters
The turbine flow meter is a transducer that measures flow rate. An example of one is shown
inside a tube in Fig. V-9. The rotation speed of the propeller is proportional to the fluid flow rate
in the tube. A coil of wire rests just outside of the tube (the “pickup”). As a permanent magnet
on one of the turbine rotors moves past the coil, it induces a current pulse proportional to the
speed of the rotor via electromagnetic induction. The rate at which the current pulses occur is
equal to the rate at which the turbine is rotating. The current is transmitted to a gauge calibrated
to read the fluid flow rate of the liquid or gas.
Figure V-9. A turbine flow meter transducer.
Flow meters are an important type of transducer often used by technicians to monitor the flow
rate of a fluid. In industry, flow meters are used to measure, record and control the movement
of fluids such as air, natural gas, water, gasoline, hydraulic oil, liquid chemicals and any other
fluid material.
Anemometers
Anemometers measure wind speed. You have probably seen an anemometer like the one in Fig.
V-10 at a weather station or an airport. They come in many shapes, but often look like, and
operate in the same manner as the turbine flow meter. Some systems designed to measure larger
flow rates rely on the rotor blades turning a generator shaft. Either method of measurement
changes the rate of gas flow into a mechanical motion that can be converted into an electrical
signal. The signal provides the readout in correct units of wind speed.
169
Figure V-10. Anemometers are used to measure wind speed.
Where are fluid transducers used?
Barometers are used by the weather bureau to measure the atmospheric pressure. These
measurements are used to predict what kind of weather is coming.
Gasoline pumps have transducers. In this use, a transducer measures how much fuel you put in
your gas tanks. The transducer also gives a readout that tells you how much you must pay for
the fuel.
Many types of industrial machines and earth moving equipment are operated by hydraulic
systems. Technicians often use pressure gauges and flow meters to troubleshoot systems.
Many high volume air conditioning systems rely on fluid transducers for information. Here, the
sensor provides data about the pressure and movement of airflow from one part of the building to
another. A technician (or a computer) can use this information to open and close air ducts, or
turn the system ON or OFF.
These are only a few examples of the uses for fluid transducers. The operating condition of
almost any fluid system can be sensed with the help of a transducer. The sensor provides data
about the system. You can use the information to change the behavior of the system - if that's
what you want.
Example V-2: Turbine flow meter.
A commercial turbine flow meter is calibrated to pick up 1200 pulses per gallon. The flow meter
registers 13,200 pulses per minute. Find the flow rate and the total volume pumped in 5 minutes.
Solution: ngal = 1200 pulses/gal
f = 13,200 pulses/minutes
QV = ?
QV = f = 13,200 pulses/min = 11 gal
ngal
1200 pulses/gal
min
Total volume in 5 minutes:
V = QV∙t = (11gpm)(5 min) = 55 gallons
170
PROBLEM SET 18: FLUID FLOW TRANSDUCERS
1. A device used to "sense" the conditions of a fluid system is called a
"________________________________".
2. The turbine flow meter can measure the _______________of a liquid or gas in a pipeline.
3. A certain flow meter is to be calibrated. If a flow rate of 2 gal/min produces 25,000 pulses in
5 minutes, how many pulses correspond to one gallon of flow?
4. What is the flow rate through a pipe if a flow meter is calibrated at 1500 pulses/gallon and
registers 60,000 pulses in a 10 minute period?
5. A commercial flow meter has a maximum rating of 3000 pulses per minute. The meter is
calibrated as follows. A five-gallon pail is filled in 60 seconds. During the time interval, the
flow meter readout registers 2000 pulses.
A) Find the number of pulses per gallon for this flow meter.
B) What is the maximum flow rate the meter can measure, in gallons per minute?
171
6. A commercial turbine flow meter is calibrated at 1200 pulses per gallon. The digital readout
shows 13,000 pulses per minute.
A) Find the flow rate in gal/min.
B) Find the total volume moved in 5.0 minutes.
172
F. FLUID PRESSURE TRANSDUCERS
The Bourdon Tube
A Bourdon tube is a fluid transducer designed to detect pressure. It converts fluid pressure into a
mechanical movement. The Bourdon C-tube is shown below in Fig. V-11. Fluid pressurizes the
hollow C- shaped tube through one end, while the other end is sealed. Pressure is force per unit
area. The pressure is constant on both the inner and outer walls, but the outer wall has more
surface area and thus experiences a greater force. Higher pressures make the C-tube straighten
out, causing the attached needle to move up in proportion to the change in pressure. Lower
pressures cause the Bourdon tube to coil up, moving the needle downward. Fluid pressure
changes are converted into a mechanical change in position.
Figure V-11. A C-shaped Bourdon tube.
Figure V-12. A spiral shaped Bourdon tube.
A spiral shaped Bourdon tube, as in Fig. V-12, provides more change in surface area and thus a
more sensitive pressure measurement. The Bourdon C-tube is most often used in pressure
gauges. It is extremely rugged and can be moved "as is" to a repair site, or mounted on a panel.
Its mechanical gear system can even be connected to a variable resistor, instead of a gauge
needle. A meter can measure the voltage across the resistor, current flow through the resistor, or
changing resistance in the resistor. This measured value is proportional to the amount of
pressure applied to the input side of the Bourdon tube. The metal in the tube experiences fatigue
over time and must be periodically recalibrated.
It is common practice to rate gauge error as a percentage of the full-scale value. For instance, if
the gauge is rated " ±3% of full scale", this means that any reading taken will be within (0.03 x
full-scale reading) on either side of the true pressure.
Barometers
The barometer is a transducer designed to measure the pressure of the air that surrounds it. Its
output signal can be either mechanical or electrical.
The most common type of barometer in use today is called the "aneroid barometer". The word
"aneroid" means "without air". That describes a part of the device pretty well. Figure V-13
shows an aneroid barometer.
173
Figure V-13. An aneroid barometer.
The heart of the aneroid barometer is capsule-shaped in appearance. The capsule is the sensor
portion and is usually made of very thin flexible metal. Part of the air inside the capsule is
removed (that's why the word aneroid is used), so that it’s at a lower pressure inside than outside.
Higher air pressure outside causes the capsule to contract. Lower air pressure outside lets the
capsule expand. The contraction and expansion of the capsule are mechanical responses to
changes in fluid pressure. A needle can be connected to the capsule to read out the changes in
pressure. Since pressure declines with altitude, aneroid barometers can be calibrated to measure
altitude based on pressure changes. In this case they are referred to as altimeters.
The readout scale of a barometer is calibrated in pressure units of millibars, inches of mercury,
inches of water or pounds per square inch. The readout of an altimeter is given in feet or meters.
In addition to the aneroid capsule, bellows and diaphragms can also be used to sense fluid
pressure. Each device produces a mechanical movement in response to a pressure change. The
mechanical movement operates a meter needle or affects the readout of an electrical
measurement device.
The Differential Pressure (“D/P”) Cell (Lab 60)
The differential pressure (D/P) cell provides a mechanical or electrical output signal when two
different pressures are compared. Keep in mind that the D/P cell measures pressure difference,
not absolute pressure. The output signal may show up as the movement of a needle or appear as
numbers on a digital display. Some d/p cells provide data to a computer, which in turn controls a
process or machine.
The configuration in Fig. V-14 utilizes a flexible diaphragm with either a strain gauge or a
piezoelectric crystal attached to it. A pressure difference across the diaphragm (P = P2-P1)
produces a change in electrical resistance in the strain gauge or a change in voltage difference in
the piezoelectric crystal proportional to the pressure difference. It will even tell you the sign of
the pressure difference. The flexible diaphragm can also be connected to a variable resistor or a
mechanical needle to measure the pressure difference.
174
Pressure
Port 1
(P1)
Pressure
Port 2
(P2)
flexible
diaphragm
mounted
strain
gauge or
piezoelectric
crystal
Figure V-14. Motion of the flexible diaphragm in this D/P cell can be measured with a strain
gauge, a piezoelectric crystal, a variable resistor or a mechanical pointer needle.
The D/P cell in Fig. V-15 measures the change in capacitance produced by a differential
pressure. The flexible, conducting diaphragm in the center responds mechanically to pressure
differences between P1 and P2. The charged capacitor plates C1 and C2 are fixed, but they do
have holes in them to allow the fluid pressure through to the center diaphragm. The changing
position of the center diaphragm changes the capacitance between it and C1, as well as it and C2.
The two changing capacitances produce a change in voltage that is a sensitive indicator of the
pressure difference across the center diaphragm.
P1
C1
flexible, conducting
diaphragm
C2
P2
Figure V-15. A capacitance type d/p cell.
In all electrical output d/p cells one of the two ports is called the "high side". The other is called
the "low side". The high side is normally designated “P1” and the low side is "P2". When the
pressure applied to the high side is greater than the pressure applied to the low side the output
signal has a positive (+) sign. If the pressure at the high side increases, the output increases. If
175
the pressure at the high side is lower than the pressure at the low side the output signal is
negative (-) sign. If the pressures at the high sides and low sides are equal the output signal is
zero.
A U-tube manometer or a d/p cell can be used to detect absolute pressure (Pabs), gauge pressure
(Pg), and the pressure difference (P). The type of pressure detected depends on the reference
pressure used. If the reference is in a near vacuum, the pressure measured is an absolute pressure
(Pabs). If the reference is atmospheric pressure, the measured pressure is a gauge pressure (Pg).
If the reference is neither vacuum nor atmospheric, then the measured pressure is a pressure
difference (P). It makes no difference which side is connected to the reference pressure in a Utube, but it does matter in the case of a d/p cell.
In Lab 60, you'll calibrate a d/p cell and produce a calibration chart. The D/P cell used in this lab
has a voltage output signal that's proportional to the pressure difference (P) across the ports on
the high side and the low side.
Often, a D/P cell is connected to other transducers to measure quantities other than pressure
difference. If the pressure difference is proportional to some other quantity, such as temperature
or flow rate, then it can be used as part of a system to measure those quantities. An orifice (a
restriction in a pipe) can be used with a D/P cell to measure the flow of a fluid. The orifice
produces a pressure difference between the upstream and downstream sides that is proportional
to the fluid velocity. D/P cells are often used to measure fluid velocity in this manner.
176
PROBLEM SET 19: FLUID PRESSURE TRANSDUCERS
1. Regardless of shape, all Bourdon tubes change pressure difference into
___________________.
2. Place a check mark next to the devices that are fluid transducers:
a.___ thermometer b.___ accelerometer
c.___ Bourdon tube d.___ barometer
e.___ radar
f.___ turbine flow meter
g.___ strain gauge
3. If a bellows moves 0.5 inch when the applied pressure is 10 psi, how much pressure is
required to move the bellows 1 inch?
4. What is the total force applied to a diaphragm if the area of the diaphragm is 0.1 in2 and the
fluid pressure is 0.3 psi?
5. How much fluid energy is required to change the volume of an element by 0.001 in3 if the
pressure causing the volume change is 10 psi?
6. A force of 5 N is applied to the fixed end of a bellows that has a cross-sectional area of 0.5 m2.
What pressure is exerted on the fixed end of this bellows?
7. Place a check mark next to the units you might expect to see on the gauge or meter of a
Bourdon tube or aneroid barometer.
a.___ ohms
b.___ mm of mercury
c.___feet of H2O
d.___psi
e.___ volts
f.___millibars
177
8. A d/p cell is connected to a hand pressure pump on the high pressure side and is open to the
atmosphere on the low pressure side. When the pump gauge reads 2 psi, the DMM voltage
output of the d/p cell is 10 mV. When the gauge reads 4 psi, the DMM reads 20 mV. At a gauge
reading of 10 psi, the DMM reads 50 mV.
A) Is the relationship between P and voltage linear? Explain your reasoning.
B) What P value is indicated by a DMM reading of 32 mV?
C) If the P value is 12 psi, what will the DMM reading be?
D) In this case, does the “P” reading indicate gauge pressure or absolute pressure?
Explain your reasoning.
9. If the same d/p cell described above is used, and the DMM reading is -10 mV (a negative
voltage), what is the gauge pressure of the input side?
178
LAB 60: THE DIFFERENTIAL PRESSURE CELL
OVERVIEW
You have learned how to measure pressure using a manometer and a gauge. In this lab, you will
use a differential pressure cell (d/p cell) to measure gas pressure.
A differential pressure cell is a transducer used to measure the pressure between two regions.
These regions may contain gases or liquids. This transducer converts pressure to an electrical
output. Such output may be used in an electronic control system or used to display readout to an
operator. The d/p cell output is often used to feed data to a computer about the state of liquid or
gaseous substances in complex systems. By using such a transducer, an operator does not have
to stay close to the system being measured. This is important when working with dangerous
substances or in harsh environments.
A differential pressure cell is one of a family of pressure cells. The word differential implies that
the device measures pressure between two points. Pressure transducers may also be used to
measure gauge pressure, or absolute pressure.
In a d/p cell we measure the difference in pressure between two points. Unlike a U-tube
manometer, in a d/p cell it is important to know which of the two points is at higher pressure.
Correctly connecting the d/p cell guarantees that the output signal is a positive voltage. The side
at higher pressure is called the high side (“P1”). The side at lower pressure is called the low side
(“P2”).
Differential pressure cells use different methods to measure pressure. However, the general idea
is that a physical change in the shape of the cell sensor causes an electrical signal to be created.
This signal is proportional to the pressure difference. The signal is amplified and can be
measured using conventional measuring instruments. You will make these measurements in this
lab. You will also compare the readings of the d/p cell to those of a compound gauge. By
making a few different measurements, you will be able to calibrate the d/p cell.
Calculations for Lab 60:
For data points 1 - 19 find differential pressure:
P = P2-P1
where P2 is read off the 30-0-30 gauge and P1 is read off the 0 - 15 psi gauge. Label the graph
on the analysis page, using units of psi on the vertical axis and mV on the horizontal axis. Label
the intersection of the axes as 0 mV and 0 psi, respectively.
Plot the graph using the results recorded in Data Table 1. Draw the straight line that best
approximates the data points.
What does it mean to have a negative voltage on the DMM? In the space below the graph on the
analysis page, write an explanation of your observations from data points 20 and 21.
179
LAB 60
THE DIFFERENTIAL PRESSURE CELL
OBJECTIVES:
SKETCH:
DATE____________
TABLE 1: Electrical Output of the d/p cell
#
Compound Gauge
Pressure
P2 (psi)
0-15 psi Gauge
Pressure
P1 (psi)
1
0
0
2
3
2
4
0
0
4
5
0
5
6
0
6
7
0
7
8
0
8
9
0
9
10
10
12
0
0
11
12
2
12
13
12
12
4
5
14
12
6
15
16
12
12
7
8
17
18
12
12
9
10
19
12
12
#
Differential
Pressure
P (psi)
Vacuum Pressure
P (in. Hg)
DMM Voltage
V (mV)
DMM Voltage
V (mV)
20
21
180
LAB 60: DIFFERENTIAL PRESSURE CELL
ANALYSIS
GRAPH OF VOLTAGE (Y axis) vs. PRESSURE DIFFERENCE (X axis)
V (mV)
P (psi)
QUESTIONS
1. Explain how you calibrated the d/p cell.
2. What is the purpose of opening the second pressure regulator when running this experiment?
3. Explain how the d/p cell is measuring "differential" pressure.
4. A U-tube manometer is a very accurate and sensitive measuring device. Describe a situation
where a d/p cell would be preferred over a manometer.
5. A compound gauge is not as sensitive as a U-tube manometer or a d/p cell. Describe a
situation where a compound gauge would still be preferred.
181
G. ELECTRICAL TRANSDUCERS
Introduction
An electrical transducer changes electrical input into a mechanical or electrical output.
Mechanical output signals are read on an analog scale by a needle responding to the input signal.
Electrical output signals are read as numbers on a digital display. A moving coil transducer
changes an input electrical signal into a mechanical output signal. An example of a moving coil
configuration is shown in Fig. V-16.
30
45
60
15
0
75
Indicator
Needle
Permanent
Magnet
Moving Coil
wrapped around
a spring-loaded
ferromagnetic cylinder
90
Figure V-16. Schematic drawing of a moving coil
The horseshoe magnet provides a permanent magnetic field. The moving coil, iron core, spring
and pointer are all one unit that rests between the magnets. When a current flows through the
moving coil, north and south poles form in the iron core. The larger the current, the stronger
these poles are. Since like poles repel and opposite poles attract, the moving coil moves due to
the interaction with the permanent magnet's north and south poles. As the current through the
moving coil wire increases, the coil moves, winding up the spring. At the same time the pointer
moves along a scale, producing an output reading. Coil movement stops when the force of
interaction of the magnets is counterbalanced by the opposing force of the now coiled spring.
The amount of current in the coil controls how far the pointer moves along the scale markings.
The amount of scale deflection, that is, the meter reading, is always proportional to the amount
of current flowing through the coil. While current flow is measured, it can represent amperes,
volts, ohms, pounds, revolutions per minute, pressure, gallons per minute or temperature,
depending upon what is being measured and how the scale reading is calibrated.
The Ammeter
A schematic of an analog ammeter using a moving coil is shown in Fig. V-17, connected to a
sample circuit. The analog ammeter has a “shunt” resistor connected in parallel to the moving
coil, providing two flow paths for the current from the circuit. The resistance of the shunt
(Rshunt) is much lower than the resistance of the moving coil (RMC), so that nearly all of the
current flowing through the ammeter will flow through the shunt. The voltage drop (Vshunt)
across the shunt must be the same as the voltage drop across the moving coil (VMC), since
they’re in parallel:
182
Vshunt  VMC
According to Ohm’s Law, V=IR, therefore
I shunt R shunt  I MCR MC
Nearly all of the current flowing through the ammeter is flowing through the shunt, therefore
I total  I shunt  I MC
R MC
R shunt
V-6
Both resistances are known in an ammeter, as is the full scale current reading for the moving
coil, which is typically limited to about 50 A to protect the delicate meter components. The
total current is then proportional to the full scale current determined by IMC, RMC and Rshunt. The
scale switch on the ammeter allows you to connect to a different value of Rshunt, depending on
what you want your full scale current limit to be. Ammeters must be connected in series to
measure the full current flow. If they are connected in parallel, almost all of the current will be
diverted from the circuit to the ammeter, resulting in an inaccurate reading and, probably, a
blown shunt resistor fuse.
Indicator
Needle
Moving Coil
with wire resistance
RMC
Permanent
Magnet
Rshunt
Analog Ammeter
Ammeter Probes
+
Sample Circuit
Rload
Figure V-17. Schematic of an ammeter attached to a sample circuit.
183
Example V-3. Current through an ammeter.
In the ammeter above, the maximum current through the moving coil (for full scale deflection) is
50 A. The moving coil resistance is 1  and the shunt resistance is 0.01 .
a) What current through the ammeter will produce full scale deflection of the needle?
IMC = 50 x 10-6 A
RMC = 1 
Rshunt = 0.01 
Itotal ≈ Ishunt = ?
R
 1 
-4
I total  I shunt  I MC MC  50 x 10-6 A 
  50 x 10 A  5 mA
R shunt
0.01



b) What is the current reading if the needle is only pointing to half of the full scale value?


Itotal = ½ Ifull scale = ½ 5 mA = 2.5 mA
The Voltmeter
The analog voltmeter is shown in Fig. V-18. It has a very large resistor (Rmeter) connected in
series to the moving coil assembly. Its value can range from hundreds of thousands to millions
of ohms. This resistor is much, much larger than that of the moving coil, and since they are in
series, the total resistance of the voltmeter is pretty much Rmeter. The voltmeter is connected in
parallel with the circuit to be measured, so that the voltage drop across the meter is the same as
the drop across the circuit load. Using Ohm’s Law, we can calculate the voltage drop given
Rmeter and IMC:
Vcircuit  Vmeter  I MCR meter
V-7
The value of Rmeter is so big that if the voltmeter is accidentally connected to the circuit in series,
the total current in the circuit will drop precipitously. When the voltmeter is connected in
parallel, the large value of Rmeter insures that only a tiny fraction of the circuit current will flow
through the meter. As in the ammeter, the current through the voltmeter coil is limited to about
50 A.
Indicator
Needle
Moving
Coil
Permanent
Magnet
Analog Voltmeter
Rload
+
-
Sample Circuit
Figure V-18. Analog Voltmeter circuit.
184
Rmeter
Example V-4. Measuring Voltage with an analog voltmeter.
The analog voltmeter above has a 200,000  resistor. If the full scale moving coil current is 50
A, and the needle is pointed at one quarter of full scale, what is the measured voltage drop in
the circuit?
Rmeter = 2.0 x 105 
IMC (full scale) = 50 A
V = ?
V = ¼ [IMC (full scale)]Rmeter = (¼) (50 x 10-6 A) (2.0 x 105 ) = 2.5 V
Digital Multimeter
A Digital Multimeter (DMM), has an analog-to-digital converter that changes the meter current
electronically to a digital output reading instead of mechanically moving a pointer.
185
PROBLEM SET 20: ELECTRICAL TRANSDUCERS
1. A moving coil transducer voltmeter has a 100,000  resistor in series with the moving-coil.
Full scale deflection of the meter occurs at 50 µA. Find the measured voltages for each of the
following meter deflections: zero, 1/4 scale, 1/2 scale, 3/4 scale and full scale. How might
these values be used?
2. A voltmeter to be calibrated has a 150,000  resistor. The moving-coil element can take no
more than 100 µA current for full scale deflection. What is the voltage reading for full scale
deflection of the pointer?
3. An ammeter has shunt resistances of a) 0.0001 , b) 0.0005  and c) 0.001 . The moving
coil resistance is 1 , and the full scale moving coil current is 50 A. What are the full scale
total currents corresponding to each of these shunt resistances?
186
The Electrostrictive Transducer
The dimensions of certain crystals change when they are exposed to an electric field. This is
called the electrostrictive effect. It is opposite in effect to the piezoelectric effect. With the
piezoelectric effect, electrical charge develops on the surface of a crystal when the crystal is
placed under strain. With the electrostrictive effect an applied voltage on a crystal causes it to
change shape and size. Changing size also causes a change in resistance of the crystal.
When a voltage is applied the crystal will change shape depending on the amount of voltage.
The external power source gives a small current flow. As the crystal changes size its resistance
changes as well. From Ohm's Law we know that I = V/R. Because of this, changing the
resistance also changes the current through the crystal. Electrostrictive crystals are commonly
used to sense and measure the strength of electric fields around transformers and power lines.
When the voltage source across the electrostrictive crystal is varied rapidly, like an AC signal,
the crystal changes its shape rapidly. This can cause crystal vibrations that, in turn, produce
sound waves. Some electrostrictive transducers are used to produce sound waves of accurately
known frequency. These are found in sonar. The word sonar is shorthand for SOund NAvigation
and Ranging. Sonar is used for a number of applications: navigation, locating objects
underwater, or other ships. Sonar can be divided into two subsystems, one that transmits sound
energy and a second subsystem to receive sound energy from returning echoes or other sound
sources. The transmitter uses a vibration crystal transducer (electrostrictive effect) to change
electrical signals into sound waves. The receiver uses a microphone (piezoelectric effect) to turn
the incoming sound waves into electrical signals. Figure V-19 shows an example of a submarine
using sonar to find a submerged UFO in the Bermuda triangle.
d
Figure V-19. Submarine sonar.
The transmitter on the submarine uses its vibrating crystal transducer to generate sound waves of
a given frequency. These sound waves move outward from the submarine until they bounce off
the UFO. Some of the waves that bounce off the UFO are reflected back to the receiver on the
submarine. Here the piezoelectric transducer converts the echo waves into electrical signals that
are fed into a computer. The computer uses the equation d=v·t/2 to determine the distance to the
UFO (since the distance to the UFO is only half of the round trip). This is an easy calculation
since the time from transmission of the initial sound wave to the time the reflected echo returns
is measured and the speed of sound is known.
187
On the transmitter a high frequency AC voltage is applied to the crystal, causing it to rapidly
change shape and move mechanically. These movements create sound waves that move out into
the water. The receiver transducer, making use of the piezoelectric effect, responds to the sound
waves - successive regions of high and low pressure - striking the crystal, by deforming in
response to the strain. This strain causes a voltage signal to be produced and sent to the
computer.
The Photoconductive Transducer
The electrical resistance of some materials changes when they are exposed to light of the right
wavelength. Photoconductive transducers are made from these types of materials.
Most photoconductive cells are made of a silicon material. Silicon has an unusual property; in
the dark, it has a very high resistance, but when light strikes it, its resistance is reduced. The
decrease in the resistance is proportional to the amount of incident light. Therefore, the incident
light is converted to an output electrical signal and can be used as a transducer to measure input
light levels.
Other materials respond to light in a manner similar to silicon and are used as transducers. These
include cadmium-sulfide and cadmium-selenide. In the absence of light both have very high
resistance, but when light falls on them their resistance decreases rapidly.
The photoconductive circuit might direct the shutter opening or aperture of a camera. Current
flows in response to the change in resistance of the transducer, which is controlled by the amount
of light received. The shutter then will open more for less light and less for a higher level of
light.
188
PROBLEM SET 21: ELECTRICAL TRANSDUCERS
1. A fish finder used by commercial deep-sea fishermen locates fish in water up to 160 meters
deep. An electrostrictive transducer is the main device in the system and it will produce a
pressure pulse of 1 psi when a short-duration voltage pulse of 300 mV is applied across it.
A. Find the voltage required to produce an 8.3 psi pressure pulse, assuming the voltage versus
pressure is a linear relationship.
B. How far down below the boat is a school of fish located if the piezoelectric detector is
activated 0.0726 sec after it was transmitted? The sound-pressure wave is traveling at 1460
m/s.
2. The transmitter on a submarine sends out a sonar signal. Four-tenths of a second later an echo
from an underwater contact is detected by the receiver on the submarine. How far is the
contact from the submarine if the speed of sound in seawater is 1460 m/s?
189
3. When exposed to light, a certain type of silicon changes its resistance. The name of the device
using this type of silicon is
_______________________________________________.
4. A photovoltaic cell used to turn streetlights off at dawn produces 0.33 V when illuminated by
10 W/m2 of light radiation. A current of 2.2 mA is delivered to a 100  resistor in series with
the cell at that light level. Find the internal resistance of the cell, Rint. Note that R = Rint +
Rload where Rload = 100 . Use Ohm's Law and solve for Rint.
190
H. LINEAR VARIABLE-DIFFERENTIAL TRANSFORMER (LVDT)
The LVDT is a position or displacement sensor. It is used to find out where something is, or
how far something has moved, or whether or not there is a difference in length between two
objects.
The input signal of an LVDT is mechanical. A mechanical movement of a shaft produces an
electrical output signal. The LVDT transformer produces the output signal.
Figure V-20 shows one type of LVDT. It consists of a magnetic core, a group of coils and
appropriate circuitry, which is also shown in the figure.
AC (Vin)
Primary Coil
Sliding
Ferromagnetic
Core
Vout
Secondary Coils
Figure V-20. LVDT schematic with Vout = 0.
AC current flowing in the primary coil produces an oscillating magnetic field which induces an
opposing voltage between the left and right secondary coils. When the core is in the center of
the secondary coils the voltages in the secondary coils cancel each other out, as shown in Fig. V20, and the output voltage (Vout) is zero. In Figure V-21, the core has been moved to the right,
leaving some of the left secondary coil surrounding nothing but air.
AC (Vin)
Primary Coil
Sliding
Ferromagnetic
Core
Vout
Secondary Coils
Figure V-21. When the core slides to the right, the induced voltage is lower in the left secondary
coil than in the right, creating a measurable Vout.
191
The primary fluctuating magnetic field is more poorly transmitted to the left secondary coil in
this case, and a measurable Vout will be produced, with the right side voltage higher than the left
side. The value of Vout is proportional to the distance the core has been moved. If the core is
moved to the left by the same amount, an equal Vout will be produced, but with the opposite
polarity. Any movement of the core from its centered position results in a net voltage output.
The LVDT is designed so that the voltage output has its maximum positive value when the core
is all the way out.
192
LAB 62: THE LINEAR VARIABLE DIFFERENTIAL TRANSFORMER
OVERVIEW
The linear variable differential transformer (LVDT) is a device designed to measure short linear
distances. These instruments are extremely accurate. They are used in industry to sense motion
of an object, to control movement of components, and to measure the travel of moving parts.
How does the LVDT work? When the LVDT is connected to an object, its position causes an
electrical signal to be generated. The output voltage changes with the position and can be
calibrated to measure distances.
LVDT’s are used to measure all kinds of motion. For example, an LVDT can be used to
measure a robot's motion. It can measure the depth that a probe enters into a test area (known as
a gauge probe when used this way). It can measure the distance a drill bore has moved. There
are many more such uses for this versatile tool. You will consider some of these in the questions
section of this lab.
In this lab you will calibrate the LVDT by finding the relationship between plunger stroke and
output signal. You will then use the calibrated LVDT to measure the number of screw turns per
unit length (screw pitch).
You will plot the data points from Table 1. Draw the straight line that best approximates the data
points. This line should pass through as many points as possible.
The Screw Pitch = DT/N, where DT is the distance traveled and N is the number of turns.
193
LAB 62 LINEAR VARIABLE DIFFERENTIAL TRANSFORMER (LVDT)
OBJECTIVES:
SKETCH:
Date_____
TABLE 1
Displacement from
Zero Point
D (mm)
LVDT Output
Voltage
V(V)
Displacement from
Zero Point
D (mm)
0
1
2
-1
-2
3
4
-3
-4
5
-5
LVDT Output
Voltage
V (V)
TABLE 2
# Screw
Turns
N
LVDT
Output
Voltage
(V)
Measured
Distance
(mm)
Distance
From
Graph
(mm)
5 (CW)
10 (CCW)
Total Measured Displacement of ten turns (DM): _________ mm
Total Displacement of ten turns from graph (DG): ________ mm
% Error: ___________ %
Screw Pitch 
DG
: __________ mm/turn
N
194
LAB 62: LINEAR VARIABLE DIFFERENTIAL TRANSFORMER (LVDT)
Voltage Output (V)
(V)
ANALYSIS
GRAPH OF Voltage (Y axis) vs. Displacement (X axis)
-6
-5
-4
-3
-2
-1
0
1
2
Displacement (mm)
1. Now that you are familiar with the LVDT, list 5 uses for it.
2. Explain how you calibrated the LVDT in your own words.
195
3
4
5
6
I. THERMAL TRANSDUCERS
Thermal transducers change thermal input data into mechanical, fluid or electrical output signals.
In most cases, thermal input data is a temperature or temperature change. When the output is
mechanical, it usually involves the movement of a metal strip, as in a bimetallic element. When
the output is fluid, it shows up as the expansion of a fluid, as in a thermometer. When the output
is electrical, it usually involves a voltage readout, as in a thermocouple, or a change in resistance,
as in a thermistor.
Liquid-in-Glass Thermometer
This is probably the thermometer you are most familiar with; the one that gets stuck under your
tongue (hopefully) when you’re sick. The liquid in the glass is often mercury, but the
thermometers in your lab benches use alcohol. Thermal energy is converted into fluid kinetic
energy, causing the fluid in the glass to expand with increasing temperature. The expansion is
proportional to the change in temperature.
Bimetallic Strip
The bimetallic strip converts thermal energy into mechanical energy. It is made up of two
different metals, each in the shape of a strip, which are tightly bonded together. When the
temperature rises, each metal expands at a different rate. Since they are bonded together, they
can't slip by one another. The expansion causes the strip to bend in an arc. In Fig. V-22,
material B expands more than material A when heated, and contracts more than material A when
cooled.
A
A
A
a) room temperature
B
b) above room temperature
B
B
c) below room temperature
Figure V-22. Shapes of the bimetallic strip.
Often it is useful to wrap the strip into a coil and to secure one end. As the temperature changes,
the coil winds up a little or unwinds, depending upon which metal is on the inside. Either way,
the free end of the coiled strip moves. This movement represents a transfer of thermal
information to mechanical information. Figure V-23 shows a common use of a bimetallic strip
as the main sensor in a thermostat. When the room is too cool, the bimetallic coil contracts and
the glass tube is tilted such that the (conductive) liquid mercury in it forms a contact with the
heater circuit and the furnace comes on. As the heat in the room rises, the coil expands and the
glass tube tilts more and more until the mercury flows away from the exposed wires, shutting the
heating system off. When you adjust the temperature dial you are adjusting the coil to tip at the
desired temperature. This is a very common, inexpensive and fairly efficient way to heat a
house.
196
Figure V-23. Bimetallic strip in a thermostat.
Thermocouples
In previous courses we covered the thermocouple and we have used them a number of times.
The thermocouple is a thermal transducer. A temperature difference along a wire causes
electrons to gather (or “huddle”) near the cold end, producing a voltage difference. The resulting
voltage difference is nearly proportional to the temperature difference over the range of the
specific type of thermocouple. Thermocouples actually use two different types of metal that
produce different amounts of electron transport in response to a temperature change, but the
theory is essentially the same.
Resistive Temperature Detectors (RTD’s)
In a normal conductor, temperature increases increase the random motion of electrons and hence
increase the resistance of the material. Over some range of temperature, the change in
resistance is proportional to the change in temperature. Resistive Temperature Detectors
(RTD’s) take advantage of this phenomenon. The change in resistance can be converted directly
to a temperature, generally on a digital multimeter.
Thermistors
A thermistor is a semiconductor device whose resistance changes with temperature. Normally,
the resistance of metals increases with increasing temperature, as with RTD’s. In a thermistor
however, the resistance decreases as the temperature increases. The resistance is inversely
proportional to the temperature. This is because higher temperatures weaken the bonds
between nuclei and electrons, allowing the free flow of electrons, which in turn means less
resistance. In metallic conductors, the electrons are already loosely bound and free to move.
Increasing the temperature in metals increases the random motion of the electrons. This causes
more collisions between electrons and thus results in higher resistance. Thermistors are made of
oxides of nickel, manganese, copper and other metals. .
The thermistor is the most sensitive temperature sensor in common use. Its largest disadvantage
is that it gives a linear response only over a very narrow temperature range. Thermistors are
used as variable resistors in bridge circuits and as temperature sensors. In each case it acts as a
transducer, converting a thermal input into an electrical output.
197
PROBLEM SET 22: THERMAL TRANSDUCERS
1. For the thermal transducers listed, place a check mark by those whose output signal is not
necessarily electrical:
a. ___ thermometer
b. ___ bimetallic
c. ___ thermocouple
d. ___ thermistor
2. A transducer that depends on the bimetallic strip to control heating systems is called a
_______________________________.
3. A type of transducer that has a decrease in electrical resistance when the temperature increases
is a _________________________.
4. A thermistor probe used with an oral thermometer has the following specifications:
Sensitivity: -30 mV/Cº
Linear accuracy: + or - 0.1% from 0 ºC to 1000 ºC
Resistance: 2250  at 25 ºC
Voltage: 1406 mV at 25 ºC
a. Find the current in the thermistor at 25 oC with the given specifications. Hint: Use Ohm's
Law.
b. Find the resistance of the thermistor at 37 oC. Hint: Assume current is constant while voltage
and resistance change in step as the temperature changes. First find the voltage change as the
temperature increases and then find the resistance using Ohm's Law.
198
APPENDIX A: ANSWERS TO PROBLEM SETS
PROBLEM SET 1
1. 0.248 m = 0.811 ft.
2. a) T = 1.40 sec
b) f = 0.712 Hz
3. a) T = 0.002 sec
b) f = 500 Hz
4. a) T = 1.27 sec
b) f = 0.788 Hz
PROBLEM SET 2
1. harmonic
2. transverse
3. longitudinal
4. require; do not require
5. wavelength
6. wavelength
7. amplitude
8. period
9. frequency
10. speed
11. 360
12. a) 180º
b) 90º
13. a) f = 92.5 MHz
b) T = 1.08 x 10-8 sec
c)  = 3.24 m
199
PROBLEM SET 3
1.
displacement from equilibrium
2
SUM OF THE
TWO WAVES
1
Phase Angle
(degrees)
0
0
90
180
270
360
450
540
630
720
-1
-2
2.
displacement from equilibrium
2
SUM OF THE
TWO WAVES
1
Phase Angle
(degrees)
0
0
90
180
270
360
-1
-2
200
450
540
630
720
PROBLEM SET 3 (CONT’D)
3. 1st harmonic:  = 8 m, f = 1.25 Hz
2nd harmonic:  = 4 m, f = 2.5 Hz
3rd harmonic:  = 2.67 m, f = 3.75 Hz
4.  = 30 cm = 0.30 m, f = 1,100 Hz
PROBLEM SET 4
1. a) joule (3); b) wavelength (1); c) x-rays (5); d) ultraviolet (2); e) Planck’s constant (4)
(c and d can be either x-rays or ultraviolet)
2. violet
3. red
4. wavelength decreases and frequency increases
5. a) 7000 Å
b) 4000 Å
c) 5550 Å
6. 3 x 1012 Hz and 3 x 1014 Hz, respectively
7. a) f = 9.23 x 107 Hz = 92.3 MHz
b) E = 6.12 x 10-26 J
8.  = 3.67 x 10-7 m
9.  = 3.98 x 10-7 m; f = 7.55 x 1014 Hz
10. E = 2.94 x 10-18 J
PROBLEM SET 5
P
1.   , where  is the irradiance in W/m2 or mW/cm2, P is the power (W or mW) and A is the
A
area (m2 or cm2)
Radiant Energy  J

2. Radiant Power 
  W
Time
s

3
2
3.  = 5.09 x 10 W/cm
4. a) P = 1 x 106 W
b)  = 5.66 x 1015 W/m2
5. t = 1.5 x 10-4 sec
6. The new power would be half of that in problem #5 (50,000 W)
7.  = 1.5 x 106 W/cm2
8. P = 50 W
PROBLEM SET 6
1. a) Z = 13; A = 27; N = 14
b) Z = 38; A = 88; N = 50
c) Z = 4; A = 9; N = 5
d) Z = 79; A = 197; N = 118
e) Z = 6; A = 12; N = 6
2. a) Ne, Ar, Kr, Xe, Rn (neon, argon, krypton, xenon, radon)
b) Al, Ga, In, Tl (aluminum, gallium, indium, thallium)
201
PROBLEM SET 7
1. 11Na24: sodium; Z = 11; A = 24; N = 13
210
82Pb : lead; Z = 82; A = 210; N = 128
2. E = 2.7 x 108 J
3. In a chemical compound the nuclei are separate from each other; they just share electrons
PROBLEM SET 8
1. c)
2. b), c) and d)
3. c) (it’s a photon, so it has no rest mass)
4.
Rest Mass
charge
penetrability
1 ____________
1 __________
1 ___________
2 ____________
2 __________
2 ___________
3 ____________
3 ___________ 3 ___________
5. c)
6. t = 6,400 years
7. P = 4.09 mW
8. T1/2 = 0.682 days (or about 16 hrs and 23 minutes)
PROBLEM SET 9
1. divided
2. power
3. A and B
4. Ein = 1563 J
5.  = 0.345 = 34.5%
6. Pout = 0.1875 hp
7.  = 0.667 = 66.7%
PROBLEM SET 10
1. Ptotal = 177 W
2. P = 452 W
PROBLEM SET 11
1. 247 panels are needed
2. P/A = 0.252 W/cm2
3. Pout = 21.8 W
PROBLEM SET 12
1. a. Qv = 60 ft3/sec
b. w = 4.5 lb
c. m = 0.140 slug
d. Ek = 113 ft·lb
e. E = 152 J
f. I = 1.27 A
202
PROBLEM SET 13
1. E = power x time = P x t
2. H = mcT
3. EE = 7.2 x 104 J = 1.72 x 104 cal
4. T = 8.6 C0
5. t = 9.30 min = 558 sec
6.  = 0.336 = 33.6%
7. t = 16.8 sec
PROBLEM SET 14
1.  = 0.20 = 20%
2. Ein = H = 0.0354 cal = 0.148 J
PROBLEM SET 15
1.  = 0.228 = 22.8%
2. Pout = 0.313 hp
3. t = 21 minutes
4. A. H = 425 cal = 1.78 x 103 J
B. Pmax = Pout = 5.93 watts
5. Overall efficiency =  = 0.269 = 26.9%; w = 14.0 lb
6. v = 6.6 m/s
PROBLEM SET 16
1. a) R = 360 
b)  = 11,300 cm = 113. m
2. The new resistance will increase to (105%/45%)Rold, so Rnew = 839 
3. R = 22.5 
4.  = 0.0126 ft. = 0.151 in.
PROBLEM SET 17
1. transducer
2. piezoelectric effect
3. V = 900 mV = 0.9 V
4. P = 2.42 x 105 N/m2
5….. pressure is applied to it.
6. a = 39.3 g’s = 384.9 m/s2 or 1.257 x 103 ft/s2
7. V = 31.2 mV
8. Sg = 66.7 mV/g
9. V = 170 mV
PROBLEM SET 18
1. fluid transducer
2. volume flow rate
3. 2,500 pulses
4. Qv = 4 gpm
5. A. 400 pulses/gal B. 7.5 gpm
6. A. Qv = 10.8 gpm B. V = 54.2 gal
203
PROBLEM SET 19
1. …mechanical movement or electrical output signals.
2. c) Bourdon tube d) barometer f) turbine flowmeter
3. P = 20 psi
4. F = 0.03 lb
5. W = 8.33 x 10-4 ft·lb
6. P = 10 N/m2
7. b, c, d, f
8. a) Yes; S = 5 mV/psi in all three cases
b) P = 6.4 psi
c) V = 60 mV
d) gauge pressure because you’re measuring the difference between one port and atmospheric
pressure.
9. P = -2 psig
PROBLEM SET 20
1. at 0: V = 0; at ¼ : V = 1.25 V; at ½ : V = 2.5 V; at ¾ : V = 3.75 V; at 1: V = 5 V
2. V = 15 V
3. a) 0.5 A; b) 0.1 A; c) 0.05 A
PROBLEM SET 21
1. A. V = 2.49 V B. d = 53.0 m
2. d = 292 m
3. photoconductive transducer
4. Rint = 50 
PROBLEM SET 22
1. a and b
2. thermostat
3. thermistor
4. A. I = 625 A B. R = 1674 
204
APPENDIX B: OSCILLOSCOPES
Relationship Between AC Current and Frequency
Frequency is a measure of how often events repeat themselves; it is therefore a rate. Events that can
be described in terms of frequency include the ticking of a clock, the position of a point on a
spinning wheel, the motion of a planet around a star, and alternating current. In the United States
commercial power companies generate 60 Hz AC current while much of the rest of the world
generates 50 Hz AC current. The term hertz means cycles per second (1 Hz = 1 cycle/s). The
waveform for a 60 Hz current is shown in Fig. II-6.
I
Imax
0
•
•
.0167
•
.0333
t (sec)
-Imax
Figure II-6. 60 Hz alternating current.
Relationship Between Frequency and Period
Frequency (f) is a measure of how many identical events or cycles take place in one second.
Period (T) is the length of time for just one of the identical events or cycles to take place. Period is
the inverse of frequency. For example, if 10 cycles occur in 1 sec, the frequency is 10 Hz. The
period is the time for one cycle, which in this case is 0.1 sec. Frequency (f) and period (T) are
related by the following formula:
f =
1
T
T =
1
f
or
Reading an Oscilloscope
When troubleshooting an electrical circuit, a technician usually uses an oscilloscope to see the
electrical signal and analyze its properties. The oscilloscope consists of a cathode ray tube, which is
marked off in grid lines and control settings. The vertical scale measures voltage and the horizontal
scale measures time. The trace on the screen can be used to measure the signal’s voltage and period
if you know the control settings. Fig. II-7 shows the “volts-per-division” (vertical) and the “timeper-division” (horizontal) controls. The figure also shows a triangle wave on the screen.
205
TIME/DIV
ms
5 2
1 0.5
10
0.2
0.1
50
20
10
5
2
1
20
s
50
0.1
0.2
0.5
1
2
.2
s
.5
HORIZONTAL
VOLTS/DIV
0.1
0.2
0.05
0.02
0.5
0.01
1
0.005
2
0.002
5
0.001
10
0.0005
VERTICAL
Oscilloscope controls and screen
The vertical control (“VOLTS/DIV”) adjusts the height of the signal displayed. Turning the
control to a higher number causes the height of the signal to decrease and vice versa. The total
height of the wave – its span from top to bottom – is called the “peak-to-peak voltage” (Vp-p). This
can be confusing since you’re really measuring from peak to trough in the vertical direction, but
somehow it has become a standard term. The peak-to-peak voltage is the number of vertical
divisions multiplied by the “voltage-per-division” scale:
 Volts 
Vp-p  # Div

 DIV 
The horizontal control adjusts the spacing between waveforms along the horizontal axis. Fig. II-8
shows how changing the time per division setting affects the screen display. The horizontal
distance between the grid lines represents elapsed time. The control setting indicates the value of
the distance between grid lines. The horizontal control has three ranges of time: seconds,
milliseconds and microseconds. In Fig. II-7 the horizontal control is set to .1 msec/div
(“milliseconds per division”). There are 1.25 cycles (5/4 of a cycle) on the screen. In Fig. II-8 the
time setting is changed to 50 sec /div. Now there are 2.5 cycles on the screen. Notice also that the
height of the signal remained the same since no changes were made to the vertical control.
206
TIME/DIV
ms
10
1 0.5
5 2
0.2
0.1
50
20
10
5
2
1
20
s
50
0.1
0.2
0.5
1
2
.2
.5
s
HORIZONTAL
VOLTS/DIV
0.1
0.2
0.05 0.02
0.5
0.01
1
0.005
2
0.002
5
0.001
10
0.0005
VERTICAL
Changing the horizontal control setting.
The horizontal distance for one complete cycle of the wave is the period (T). The period is the
number of divisions for one complete cycle multiplied by the “time per division” scale:
 Time 
T  # Div

 DIV 
The frequency of the wave is the inverse of the period.
In the following Example we analyze a triangular wave input signal. Work your way through the
example problem, noting the control settings.
207
Example: Reading an Oscilloscope.
For the cathode ray tube (CRT) oscilloscope display shown below, find a) the period (T) of one
cycle of the waveform, b) the peak-to-peak voltage of the signal, and c), the type of waveform.
TIME/DIV
ms
5 2
1 0.5
10
0.2
0.1
50
20
10
5
2
1
20
s
50
0.1
0.2
0.5
1
2
.2
.5
s
HORIZONTAL
VOLTS/DIV
0.1
0.2
0.05
0.02
0.5
0.01
1
0.005
2
0.002
5
0.001
10
0.0005
VERTICAL
Solution: a) The “time/div” control is set at 20 sec = 20 x 10-6 s per division (the Greek letter 
(“mu”) stands for “micro”). Counting the number of divisions for one period from point A to point
B on the grid face, you get 4 divisions. Thus the period (T) is:
T
 time 
 sec 
# divisions = 20
4 div = 80 sec
division
div 
b) The peak-to-peak voltage of this signal is the difference between the top-most point on the
vertical (voltage) scale and the bottom-most point. The “volts/div” knob is set at 0.5 volts/div.
Counting the number of divisions from the maximum positive trace point (top) to the maximum
negative trace point (bottom) yields 8 divisions. The peak-to-peak voltage is thus
Vp-p =
volts 
 # volts 

# divisions = 0.5
8 div = 4.0 V(peak-to-peak)
division
div 
c) The waveform is triangular.
Other common waveforms are sine waves and square waves.
NOTE: The oscilloscope probe has a “x10” switch for reducing large voltage inputs. Always
make sure the probe is set for “x1”.
208
APPENDIX C: 5-STEP METHOD OF SOLVING PHYSICS PROBLEMS
1. READ THE ENTIRE PROBLEM CAREFULLY, AND MAKE A SKETCH.
Read through the problem completely before you start to write anything down. A sketch of the
physical situation will help to clarify the ideas of the problem. If you can’t visualize the
situation, you might be missing some important concepts.
2. LIST THE GIVEN INFORMATION AND IDENTIFY THE UNKNOWN QUANTITY
ASKED FOR IN THE PROBLEM.
Write down each magnitude (number & unit) that is given, and identify it with the appropriate
letter. For example, a force of 6.50 pounds is listed as F = 6.50 lb. It is important to use the
letter that will appear in equations. For example, if a problem asks you to find how long it takes
for an event to occur, you would write “t = ?”.
3. FROM YOUR LIST OF EQUATIONS, SELECT THE EQUATION THAT RELATES
THE UNKNOWN QUANTITY TO THE GIVEN INFORMATION. REWRITE THE
EQUATION, IF NECESSARY, TO SOLVE FOR THE UNKNOWN QUANTITY
For example, if you know the density () and the volume (V), and you wish to find the mass of
an object (m=?) you would choose the equation that contains the letters , V, and m:  = m/V.
The unknown should appear by itself on the left of the equal sign in your working equation. In
the example above, where the unknown was the mass, the basic equation  = m/V would be
rewritten to read: m = V. That is the working equation form.
4. SUBSTITUTE THE KNOWN INFORMATION IN THE WORKING EQUATION,
INCLUDING ALL UNITS.
5. SOLVE THE EQUATION, INDICATING THE CANCELLATION OF UNITS, and
CIRCLE YOUR ANSWER
Check your answer to see that it has correct units. For example, if you find that the weight of an
object is in units of square feet, then an error has occurred. Also, check the magnitude of your
answer; if it is obviously physically impossible, go back and look for an error. For example, if
you find the speed of a car to be 4000 mph, the answer is not reasonable!
209
APPENDIX D: EQUATIONS
WEIGHT AND MASS EQUIVALENTS
ALWAYS TRUE
AT THE EARTH'S SURFACE ONLY
1 lb (force) = 4.45 N (force)
1 slug (mass) = 14.6 kg (mass)
1 kg weighs 9.8 N or 2.2 lb
1 slug weighs 32.2 lb or 143 N
g = 9.8 m/sec2 = 9.8 N/kg
LINEAR MECHANICAL EQUATIONS
UNITS
SI
ENGLISH
W = F·D
W = work
F = applied force
D = distance moved in the direction
of the force
J (joule)
N (newton)
m (meter)
ft·lb
lb
ft
W = w·h
W = mgh
W = work to lift a weight
w = weight
h = height lifted
m = mass
g = gravitational acceleration @ sea
level
J
N
m
kg
g = 9.8 m/s2
ft·lb
lb
ft
slug
32 ft/s2
v = d/t
v = speed or velocity
d = distance traveled
t = time of travel
m/s
m
s
ft/s
ft
s
a = (vf -vi ) t
a = acceleration
m/s2
ft/s2
F = m·a
(Newton's
second law)
F = net force
m = mass
a = acceleration caused by net force
N
kg
m/s2
lb
slug
ft/s2
F = μk·N
F = friction force (sliding surfaces)
N = normal force
μk = coefficient of kinetic friction
N (newton)
N (newton)
lb
lb
F = friction force (static surfaces)
N = normal force
μs = coefficient of static friction
N (newton)
N (newton)
F = μs·N
μ is dimensionless
lb
lb
μ is
dimensionless
Ep = ½k·d2
Ep = potential energy of a stretched or
compressed spring
k = spring constant
d = length of deformation
210
J
ft·lb
N/m
m
lb/ft
ft
LINEAR MECHANICAL EQUATIONS (CONTINUED)
UNITS
SI
ENGLISH
P = W/t
P = power
W (watt) (=J/s)
ft·lb/s
P = F·v
F = applied force
v = velocity`
N
m/s
lb
ft/s
η = Wout/Win
η = Pout/Pin
η = efficiency
W = work
P = power
η is expressed as %
work units must agree
power units must agree
p = mv
p = linear momentum
kg·m/s
slug·ft/s
Imp = F·Δt
Imp = impulse
F = net force applied
Δt = time interval of applied force
N·s
N
s
lb·s
lb
s
F·Δt = Δp
impulse = change in momentum
note: kg·m/s =
N·s
slug·ft/s =
lb·s
Momentum is a vector
pf = pi
Final Momentum = Initial Momentum In all collisions
211
ROTATIONAL MECHANICAL EQUATIONS
SI
UNITS
ENGLISH
τ = F·
τ = torque
F = applied force
 = lever arm
N·m
N
m
lb-ft or ft-lb
lb
ft
W = τ·θ
W = work
τ = applied torque
θ = angle of rotation in
radians
J (joule)
N·m
ft·lb
lb·ft
ω = θ/t
ω : omega
ω = angular speed
rev/min (revolutions per minute or RPM)
rad/s (radians per second)
α = (ωf-ωi)/ t
α : alpha
α = angular or rotational
acceleration
rad/s2 (radians per second squared)
α = τnet /I
τnet = net torque acting on an
object
α = accel. caused by the
torque
N·m
ft·lb
rad/s2
rad/s2
Ek = rotational kinetic
energy
I = moment of inertia
ω = angular speed
J
ft·lb
kg·m2
rad/s
slug·ft2
rad/s
v = ω·r
( rolling without
slipping)
v = linear velocity
ω = angular velocity in rad/s
r = radius
m/s
rad/s
m
ft/s
rad/s
ft
P = τ·θ/t
P = τ·ω
P = power
τ = torque
θ = angle
ω = angular speed
W (watt)
N·m
radians
rad/s
ft·lb/s
ft·lb
radians
rad/s
η = Wout/Win
η = Pout/Pin
η = efficiency
W = work
P = power
η is expressed as %
L = I·ω
L = angular momentum
ω = angular speed
kg·m2/s
radians/s
slug·ft2/s
radians/s
Ang. Imp.= τΔt
Angular Impulse
N·m·s
lb·ft·s
ΔL = τΔt
change in angular
momentum = angular
impulse
note:
kg·m2/s =
N·m·s
slug·ft2/s = lb·ft·s
θ : theta
Ek = ½I·ω2
radian (1 rev = 2 rad)
work units must agree
power units must agree
Angular Momentum is a vector.
Lf = Li
Angular momentum is conserved in isolated systems
212
FLUID EQUATIONS
ρ = m/V
ρ : rho
ρw = w/V
P = F/A
P = ρ·g·h
P = ρw·h
or
W = P·V
QV = V/t
QV = v·A
Qm = m/t
Rdrag = Fdrag/v
(an object moving
through a fluid)
(laminar flow)
RF = ΔP/QV
(fluid flow through a
pipe) (laminar flow)
Ep = ρVgh,
Ep= ρwVh
Ek = ½ [(ρw/g)V]v2
Ek = ΔP·QV ·t
P = 1/2[(w/g)QVv2]
= 1/2[(w/g)Av3]
= 1/2[(QVv2]
= 1/2[(Av3]
v=
(2gh)1/2
ρ = mass density
m = mass
V = volume
ρw = weight density
w = weight
P = pressure
F = applied force
A = area
P = gauge pressure at a
depth h
W = work
P = pressure difference
V = volume of fluid
QV = volume flow rate
V = volume
t = time
v = velocity
A = area
Qm = mass flow rate
Rdrag = drag
resistance
Fdrag = drag force
v = speed
RF = fluid resistance
UNITS
SI
ENGLISH
g/cm3, kg/m3
slugs/ft3
(seldom used)
g
3
3
cm , m
N/m3
lb/ft3
N
lb
Pa (pascal)
lb/in2 or lb/ft2
1 Pa = 1 N/m2
J (joule)
N/m2
m3
m3/s, m3/hr,
/s
ft·lb
lb/ft2
ft3
ft3/hr
gal/min (GPM)
grams/sec
kg/s
N/(m/s)
slug/hr
N (newton)
m/s
Pa/(m3/s)
lb/(ft/s)
ΔP = pressure difference
QV = vol. flow rate
Pa=N/m2
m3/s
lb
ft/s
psi/GPM
psi/(ft3/hr)
lb/in2 (psi)
ft3 /hr
Ep = potential energy
J (joule)
ft·lb
Ek = kinetic energy
J
ft·lb
units must cancel to give joules or ftlb
P=power
W
N/m3
w=weight density
m3/s
Qv =volume flow rate
m2
A=cross sectional area
v = velocity
m/s
g=gravitational acceleration
9.8 m/s2
 = mass density
kg/m3
v = efflux velocity
m/s
h = depth at opening
m
g = gravitational constant
9.8 m/s2
Bernouilli=s Equation: Pa + (1/2)ρva2 + ρgha = Pb + (1/2)ρvb2 + ρghb
(Conservation of energy in a fluid system)
213
ft∙lb/sec
lb/ft3
ft3/sec
ft2
ft/s
32.2 ft/sec2
slug/ft3
ft/s
ft
32 ft/s2
ELECTRICAL EQUATIONS
UNITS: SI only
VT = V1 + V2 + …
VT = total potential difference for
batteries in series
VT = V1 = V2 = …
VT = total potential difference for
batteries in parallel
I = q/t
I = electric current
q = charge
A (ampere)
C (coulomb)
W = V·q
W = electrical work
V = potential difference
q = quantity of charge
I = current
t = time in seconds
J (joule)
V
C (coulomb)
W = V·I·t
moved
V (volt)
A (ampere)
s (second)
R = V/I
R = Resistance
Ω (ohm)
C = q/V
C = capacitance
F (farad)
Ep = ½qV
Ep = ½CV2
Ep = potential energy of a charged
capacitor
J (joule)
L = V(I/t)
L = inductance
V = potential difference
I = current
t = time
H (henry)
V (volt)
A (amp)
s (second)
note: 1 Henry =
V = V·s =Ω·s
A/s A
Ep = ½LI2
Ep = potential energy of an inductor
J (joule)
P = qV/t
P = VI
P = V2/R
P = I2 R
P = power
q = charge moved
V = potential difference
I = current
R = resistance
W (watt)
C (coulomb)
V (volt)
A (ampere)
Ω (ohm)
η = Pout/Pin
η = efficiency
P = power
η as a %
W (watt)
Vsec/Vprim= Nsec/Nprim
(Electrical Transformer)
V = voltage (secondary, primary)
N = number of windings
V (volts)
pure number
214
none
THERMAL EQUATIONS
UNITS
ENGLISH
SI
TC = 5/9(TF - 32o)
TF = 9/5(Tc) + 32o
ΔTC = 5/9(ΔTF)
TC = Celsius temperature
TF = Fahrenheit temperature
oC
oF
(example:
TC = 20oC)
(example:
TF = 70oF)
ΔT = temperature
change
or temperature difference
Co
Fo
ΔTF = 9/5(ΔTC)
examples:
ΔT = 5 Co
ΔT = 9 Fo
QH = H/t
QH = heat flow rate
H = heat energy
cal/s
cal (calorie)
Btu/hr
Btu (British
thermal unit)
H = m·c·ΔT
H = heat energy
m = mass of object
ΔT = change in
temperature
c = specific heat of the
material
cal
g (gram)
Btu
lbm (mass of a
1-lb weight)
Fo
Btu/(lbFo)
RT = thermal resistance
Co/kcal/hr or
Co/cal/sec
Co
kcal/hr or
cal/s
Fo/Btu/hr
cal/s
Btu/hr
cal·cm
(sec·cm2·Co)
Co
Btu·in
(hr·ft2·Fo)
Fo
cm2
cm
ft2
in
(thermal energy
needed to heat or cool
an object)
RT = ΔT/ QH
ΔT = temp. difference
QH = heat flow rate
QH = AΔT/
QH = heat flow rate across a slab
of material
 = thermal conductivity of the
material
ΔT = temperature difference
across the material
A = cross sectional area
 = thickness of the material
215
Co
cal/(gCo) or
kcal/(kgCo)
Fo
Btu/hr
FORCE TRANSFORMERS
Mechanical Advantage for all machines:
IMA = Di/Do
AMA = Fo/Fi
Ideal Mechanical Advantage = input distance  output distance
Actual Mechanical Advantage = output force  input force
Efficiency = AMA/IMA
TIME CONSTANTS: Exponential Growth and Decay
Process
Growth
Decay
General Equations:
N = Nmax (1 - e-t/τ )
N = N0 e-t/τ
Damping of a Vibration
A = A0 e-t/τ
Draining of liquid from an open tank
(approximate)
Qv = Q0 e-t/τ
Capacitor Voltage
V = Vmax (1 - e-t/τ )
V = V0 e-t/τ
Heating or cooling to ambient temperature
ΔT = (ΔT)max (1 - e-t/τ )
ΔT = (ΔT)0 e-t/τ
A = A0 e-t/τ
Radioactive Decay
Radioactive half-life calculations
Current in an RC or RL circuit
(τ = RC or τ = L/R )
A=Ao(1/2)t/T1/2
I = Imax (1 - e-t/τ )
I = I0 e-t/τ
SPEED OF SOUND IN SALT WATER:
vs  [1449+46T+(1.34-0.01T)(S-35)+0.016z] m/s,
where z is the depth in meters, T is the temperature in degrees Celsius and S is the salinity in
parts per thousand.
216
EQUATIONS FOR VIBRATIONS AND WAVES
PERIODIC MOTION
T  2
m
k
Where m = mass on end of spring
k = spring constant
T = period of oscillation

g
Where g = acceleration due to gravity
 = the length of the pendulum string
T = period of oscillation
T  2
RESONANCE OF SOUND IN TUBES
For an open tube of length , resonance will occur for the following wavelengths.
First resonance: 1 = 2 or  = 2 1
Second resonance:  2 =  or  =  2
Third resonance:  3 = (2/3) or  = (3/2) 3
Fourth resonance:  4 =  /2 or  = 2 4
The frequency is related to the wavelength by the following equation:
f
vs

Where: f = frequency of the wave in Hz
vs = velocity of the wave in m/s
 = wavelength of the sound wave in the tube in meters.
EFFECTS OF PHASE DIFFERENCE
P = (V x I) cos
Where: P = power
V = voltage
I = current
 (theta) = phase angle between the voltage and current
217
ELECTROMAGNETIC RADIATION
The energy of an electromagnetic wave:
Where:
E = hf = hc/
h = Plank's constant = 6.63 x 10-34 Js
 = wavelength (in m)
c = speed of e-m radiation in vacuum ( 3 x 108 m /sec)
f = wave frequency (cycles/sec)
c = f
Where: c = speed of e-m wave in a transparent medium (in m/sec)
 = wavelength of e-m wave (wavelength) (in m)
f = frequency of e-m wave (in Hz)
ABBREVIATIONS FOR POWERS OF TEN
Abbreviation
p
n

m
k
M
G
T
Name
pico
nano
micro
milli
kilo
mega
giga
tera
Equivalent
x 10-12
x 10-9
x 10-6
x 10-3
x 103
x 106
x 109
x 1012
218
APPENDIX E: CONVERSION FACTORS & REFERENCE TABLES
CONVERSION FACTORS
LENGTH
1 inch (in) = 2.54 centimeters (cm) = 25.4 millimeters (mm)
1 meter (m) = 100 cm = 1000 mm = 39.4 in = 3.28 ft
1 kilometer (km) = 1000 m = 0.621 mi
1 mile (mi) = 5280 ft = 1.61 km
AREA
1 m2 = 10,000 cm 2 = 1550 in2 = 10.76 ft2
1 ft2 = 144 in2 = 929 cm2
VOLUME
1 m3 = 1000 liters () = 106 cm3 = 35.3 ft3
1 cm3 = 1 milliliter (ml) = 0.001 
1 ft3 = 1728 in3 = 28.3 liters () = 7.48 gal
1 gal = 4 qts = 3.785 liters
DENSITY
mass density of water = 1 g/cm3 = 1 g/ml = 1000 kg/m3 = 1.94 slug/ft3
weight density of water = 62.4 lb/ft3 = 9800 N/m3
weight density of air at room temperature = 0.075 lb/ft3
SPEED
1 m/s = 3.28 ft/s = 2.24 mi/hr = 3.60 km/hr
1 ft/s = 0.305 m/s = 0.682 mi/hr = 1.10 km/hr
1 km/hr = 0.278 m/s = 0.913 ft/s = 0.621 mi/hr
1 mi/hr = 1.47 ft/s = 1.61 km/hr
VISCOSITY water at room temp = 2.12 x 10-5 slug/ft-sec
air at room temp = 3.75 x 10-7 slug/ft-sec
MASS
1 kilogram (kg) = 1000 grams (g) = 0.0685 slug
1 slug = 14.6 kg
FORCE
1 newton (N) = 0.225 lb = 3.60 oz
1 pound (lb) = 16 oz = 4.45 N
PRESSURE 1 pascal (Pa) = 1 N/m2
1 bar = 105 Pa
1 lb/in2 (psi) = 144 lb/ft2 (psf) = 6900 Pa
1 atmosphere (atm) = 14.7 lb/in2 (psi) = 2117 lb/ft2 (psf) = 101,325 Pa
ENERGY
1 joule (J) = 0.738 ftlb
1 ftlb = 1.36 J
1 kilocalorie (kcal or Cal) = 1000 calories (cal)
1 cal = 4.184 J = 3.97 x 10-3 BTU = 3.077 ftlb
1 BTU = 252 cal = 778 ftlb = 1054 J
POWER
1 watt = 0.73760 ftlb/s = 3.41 BTU/hr = 0.239 cal/s
1 hp = 550 ftlb/s = 746 W
1 kilowatt (kW) = 1000 W
219
TEMPERATURE DIFFERENCE (T) 9 Fo = 5 Co
NOTE: FOR TEMPERATURE CONVERSION USE
T( o F) = 32 o F +
9 o
T( C) or
5
T( o C) =
5
 T( o F) - 32 o F
9
THERMOCOUPLE EQUATIONS
Vtotal  Vref  Vreading
 T -T
Tactual  Tmin  Vtotal - Vmin  max min
 Vmax  Vmin



In our lab exercises, Vref is the voltage corresponding to the reference temperature (obtained
from the thermocouple table), Vreading is the measured voltage difference (the voltmeter reading)
and Vtotal is the total voltage difference, the value that is converted into a temperature using the
thermocouple table. Vmin and Vmax are the voltage readings in the thermocouple table just below
and just above Vtotal, respectively. Tmin and Tmax are the temperatures in the table corresponding
to Vmin and Vmax, respectively, and Tactual is the actual temperature of what you’re measuring.
ANGLE MEASUREMENTS 1 revolution = 2 radians = 360o
220
TYPE-E THERMOCOUPLE
Chromel-Constantan Reference Junction at 0 oC
0 Co
2 Co
4 Co
6 Co
8 Co
10 Co
-20 Co
-1.15 mV
-1.26 mV
-1.38 mV
-1.49 mV
-1.60 mV
-1.71 mV
-10 Co
-0.58 mV
-0.70
-0.81
-0.93
-1.04
-1.15
0 Co
0.00
-0.12
-0.23
-0.35
-0.47
-0.58
0 Co
0.00
0.12
0.23
0.35
0.47
0.59
10 Co
0.59 mV
0.71
0.83
0.95
1.07
1.19
20 Co
1.19
1.31
1.43
1.56
1.68
1.80
30 Co
1.80
1.92
2.05
2.17
2.30
2.42
40 Co
2.42
2.54
2.67
2.80
2.92
3.05
50 Co
3.05
3.17
3.30
3.43
3.56
3.68
60 Co
3.68
3.81
3.94
4.07
4.20
4.33
70 Co
4.33
4.46
4.59
4.72
4.85
4.98
80 Co
4.98
5.12
5.25
5.38
5.51
5.65
90 Co
5.65
5.78
5.91
6.05
6.18
6.32
100 Co
6.32
6.45
6.59
6.72
6.86
7.00
110 Co
7.00
7.13
7.27
7.41
7.54
7.68
120 Co
7.68
7.82
7.96
8.10
8.24
8.38
130 Co
8.38
8.52
8.66
8.80
8.94
9.08
140 Co
9.08
9.22
9.36
9.50
9.64
9.79
150 Co
9.79
9.93
10.17
10.22
10.36
10.50
160 Co
10.50
10.65
10.79
10.93
11.08
11.22
170 Co
11.22
11.37
11.51
11.66
11.80
11.95
180 Co
11.95
12.10
12.24
12.39
12.53
12.68
o
The top three rows of the table provide voltages for temperatures below 0 C. In that case, the
column headings correspond to 0 oC, -2 oC, -4 oC, and so on.
NOTE: If the reference junction is not at 0oC, the table above gives the T value. To find the measurement
temperature, add T from the table to the reference temperature.
221
MASS DENSITY (ρ) & SPECIFIC GRAVITY
SOLIDS
ρ
g/cm3
Sp.G.
no units
LIQUIDS &
GASES
ρ
g/cm3
Sp.G.
no units
Gold
19.3
19.3
Mercury
13.6
13.6
Lead
11.3
11.3
Water
1.0
1.0
Silver
10.5
10.5
Oil
0.9
0.9
Copper
8.9
8.9
Alcohol
0.8
0.8
Steel
7.8
7.8
Antifreeze
1.125 (32oF)
1.098 (77oF)
1.125 (32oF)
1.098 (77oF)
Aluminum
2.7
2.7
Air
1.29 x 10-3
1.29 x 10-3
Balsa Wood
0.3
0.3
Hydrogen
9.0 x 10-5
9.0 x 10-5
Oak Wood
0.8
0.8
Oxygen
1.43 x 10-3
1.43 x 10-3
WEIGHT DENSITY (ρw)
To find the weight density of a substance, multiply the specific gravity of the substance
by the weight density of water. The weight density (ρw) of water is 62.4 lb/ft3 = 9800 N/m3
PRESSURE CONVERSION CHART
psi
in. of H2O
in of Hg
mm of H2O
mm of
Hg
bar
mbar
1.0
27.68
2.036
703.1
51.71
0.0689
68.95
APPROXIMATE COEFFICIENTS OF FRICTION (μ)
Material
static (μs)
kinetic (μk)
wood on wood
metal on wood
leather on wood
rubber on dry concrete
rubber on wet concrete
steel on concrete
steel on steel
0.7
0.55
0.5
0.9
0.7
0.9
0.15
0.3
0.5
0.4
0.7
0.5
0.5
0.09
222
SPECIFIC HEATS (c) OF COMMON SUBSTANCES
SPECIFIC
HEAT
cal/(g·Co) or
Btu/(lb·Fo)
SUBSTANCE
SPECIFIC
HEAT
cal/(g·Co) or
Btu/(lb·Fo)
CONVERSION
FACTORS
Air
0.24
Stone (average)
0.192
1 Btu = 252 cal
Aluminum
0.22
Tin
0.055
1 kcal = 1000 cal
Brass
0.091
Water
1.00
1 cal = 3.09 ft·lb
Copper
0.093
Ice
0.50
1 Btu = 778 ft·lb
Glass
0.21
Steam
0.48
1 cal = 4.18 joules
Iron (steel)
0.115
Wood (average)
0.42
SUBSTANCE
HEAT OF FUSION AND VAPORIZATION
MELTING
POINT (oC)
BOILING
POINT (oC)
OXYGEN
-218
NITROGEN
ALCOHOL
SUBSTANCE
LATENT HEATS
Solid  Liquid
Liquid  Vapor
-183
3.3 kcal/kg
51 kcal/kg
-210
-196
6.1 kcal/kg
48 kcal/kg
-114
78
26.0 kcal/kg
204 kcal/kg
0
100
80.0 kcal/kg
540 kcal/kg
LEAD
327
1750
5.5 kcal/kg
205 kcal/kg
SILVER
961
2212
26.5 kcal/kg
563 kcal/kg
WATER
223
APPROXIMATE VALUES OF THERMAL CONDUCTIVITY ()
MATERIAL
SI
cal·cm
sec·cm2·Co
ENGLISH
Btu·in
hr·ft2·Fo
INSULATING
MATERIALS
ENGLISH
Btu·in
hr·ft2·Fo
Air
1.96 x 10-8
5.7 x 10-4
Hair Felt
0.26
Water
1.14 x 10-4
0.33
Rock Wool
0.26
Corkboard
1.03 x 10-4
0.30
Glass Wool
0.29
Celotex
1.17 x 10-4
0.34
White Pine
0.78
White Pine
2.69 x 10-4
0.78
Oak
1.02
Marble
3.79 x 10-4
1.10
Cinder Block
2 to 3
Concrete
2.07 x 10-3
6.0
Building Block
3 to 6
Glass
1.99 x 10-3
5.8
Glass
5 to 6
Steel
0.12
350
Concrete
6 to 9
Aluminum
0.48
1400
Granite
13 to 18
Copper
0.93
2700
THERMAL CONDUCTIVITY OF METALS AT 0 OC
METAL
Thermal Conductivity
METAL
Thermal Conductivity
cal·cm
hr·cm2·Co
Btu·in
hr·ft2·Fo
Aluminum
2031
1632
Iron
718
576
Copper
3450
2784
Silver
3683
2964
Gold
2736
2208
Tungsten
1566
1260
calcm
hr·cm2·Co
224
Btuin
hr·ft2·Fo
R-VALUES FOR TYPICAL INSULATION THICKNESSES
Batts or Blankets
Loose Fill (Poured-in)
R-Value
Glass Fiber
Rock Wool
Glass Fiber
Rock Wool
Cellulosic
Fiber
R-11
3.5 – 4 inches
3 inches
5 inches
4 inches
3 inches
R-19
6 – 6.5 inches
5.25 inches
8 - 9 inches
6 - 7 inches
5 inches
R-22
6.5 inches
6 inches
10 inches
7 - 8 inches
6 inches
R-30
9.5 –10.5 inches
9 inches
13 - 14 inches
10-11 inches
8 inches
17 - 18 inches
13-14 inches
10-11 inches
R-38
12 - 13 inches
10.5 inches
* Two batts or blankets required
MOMENT OF INERTIA (I) FOR COMMON SHAPES
AXIS
AXIS
m
r1

r
AXI
r
r
AXIS
AXI
d. Solid
I = 1 mr2
2
AXIS
e. Thin-walled hollow
cylinder (wheel)
I = mr2
r

g. Solid cylinder
I = 1 mr2 + 1 m2
4
12
225
AXIS
c. Annular (ring)
cylinder
I = 1 m (r12 + r22)
2
b. Slender rod.
through center
I = 1 m 2
12
a. Particle in
circular orbit
I = mr2
r2
f. Solid
I = 2 mr2
5
RESISTOR COLOR CODES
↑
COLOR
↑
↑
FIRST
SIGNIFICANT
FIGURE
SECOND
SIGNIFICANT
FIGURE
MULTIPLIER
BLACK
0
0
1
BROWN
1
1
101
RED
2
2
102
ORANGE
3
3
103
YELLOW
4
4
104
GREEN
5
5
105
BLUE
6
6
106
VIOLET
7
7
107
GRAY
8
8
108
WHITE
9
9
109
↑
TOLERANCE
RESISTIVITY OF COMMON CONDUCTORS
MATERIAL
RESISTIVITY
μΩ·cm
MATERIAL
RESISTIVITY
μΩ·cm
Glass
1020
Lead
Silicon *
106
Aluminum
2.824
Germanium *
106
Gold
2.44
Carbon
105
Copper
1.724
Nichrome Wire
112
Silver
1.59
22
*The resistivity of semiconductors - like silicon and germanium - is very sensitive to
temperature.
226
YOUNG’S MODULUS OF ELASTICITY
ENGLISH UNITS
lb/in2
SI UNITS
N/cm2
Aluminum
9.8 x 106
6.75 x 106
Brass
13.0 x 106
8.96 x 106
Copper
17.3 x 106
11.92 x 106
Cast Iron
13.0 x 106
8.96 x 106
Steel
29.0 x 106
19.98 x 106
Quartz Fiber
7.3 x 106
5.03 x 106
Glass, Crown
9.9 x 106
6.82 x 106
MATERIAL
PIEZOELECTRIC CRYSTAL PROPERTIES
k= Piezoelectric Constant
(Voltage per unit applied
pressure per unit thickness,
V·m/N)
Coupling Coefficient
(Efficiency)
Quartz
0.055
9.9%
Rochelle salt (at 300)
0.098
78%
Ammonium dihydrogen
phosphate
0.178
25%
Ethylene diamine tartrate
0.152
21.5%
Lithium sulfate
0.165
35%
Tourmaline
0.0275
Crystal
227
9.2%
LIGHT
METALS
PERIODIC TABLE OF THE ELEMENTS
IA
VIIIA
Hydrogen
Helium
1.0000
4.003
In the periodic table the elements are arranged in order of increasing atomic number.
H
NON METALS
He
Vertical columns headed by Roman. Numerals are called Groups. A horizontal
sequence is called a Period. The most active elements are at the top right and
bottom left of the table. The staggered line (Groups IIIA and VIIA) roughly separates IIIA
IIA
IVA
VA VIA VIIA
2
1
metallic from non-metallic elements.
Lithium Beryllium Group IA includes hydrogen and the alkali
Boron
Carbon
Nitrogen Oxygen
Fluorine
Neon
Groups- Elements within a group have
metals.
similar properties and contain the
6.939 9.012 Group VIIA includes the halogens.
10.811 12.01115 14.007 15.999
18.998 20.183
same number of electrons in their
outside energy shell.
elements intervening between groups IIA
Li Be The
B
C
N O
F Ne
and IIIA are called transition elements.
elements intervening between groups IIA Periods- In a given period the
3
4 The
5
6
7
8
9 10
and IIIA are called transition elements.
properties of the elements gradually
pass from a strong metallic to a strong
nonSodium Magnesium The elements intervening between groups IIA metallic nature, with the last number of Aluminum
Silicon Phosphate Sulfur
Chlorine
Argon
and IIIA are called transition elements.
a period being An inert gas.
22.990 24.312
Na Mg
11 12
Potassium Calcium
Short vertical columns without Roman
numeral headings are called sub-groups.
30.974 32.064
35.453
39.948
Al
13
Si
14
P S
15 16
Cl
17
Ar
18
Scandium
Titanium
Vanadium Chromium Manganese
40.08
44.956
47.90
50.942
51.996
K
19
Ca
20
Sc
21
Ti
22
V
23
Cr
24
Rubidium Strontium
Yttrium
Zirconium
115.17
87.62
88.905
91.22
92.906
95.94
99
Rb
37
Sr
38
Y
39
Zr
40
Nb
41
Mo
42
Tc
43
Cesium
Barium
Hafnium
Tantalum
Tungsten
Rhenium
Osmium
Iridium
Platinum
Mercury
Thallium
Lead
178.49
180.95
183.85
186.21
190.2
192.2
195.09
196.97 200.59
204.37
207.19
Ta
73
W
74
Re
75
Pt
78
Au Hg
79 80
Tl
81
Pb
82
Cs
55
Ba 57-71 Hf
56
72
Francium
Radium
22.3
(226)
Fr
87
Ra
88
Lanthanum Cerium
138.91 140.12
89-103
Cobalt
28.086
39.102
132.90 137.34
Iron
26.981
54.938 55.847 58.933
Nickel
Copper
Zinc
Gallium
Germanium
Arsenic Selenium
Bromine
krypton
58.71
63.54
65.37
69.72
72.59
74.922 78.96
79.909
83.80
Cu Zn
29 30
Ga
31
Ge
32
As Se
33 34
Br
35
Kr
36
Mn Fe Co Ni
25 26 27 28
Niobium Molybdenum Technetium Ruthenium Rhodium Palladium
101.07 102.91
106.4
Ru Rh Pd
44 45 46
Os Ir
76 77
Praseodymium Neodymium Promethium Samarium
Tin
Antimony Tellurium
Iodine
Xenon
114.82
118.69
121.75 127.60
126.90
131.30
Ag Cd
47 48
In
49
Sn
50
I
53
Xe
54
Gold
150.35
Europium Gadolinium Terbium Dysprosium Holmium
144.24
(147)
Ce
58
Pr
59
Nd
60
Pm Sm
61 62
Actinium
Thorium
Protactinium
Uranium
227
232.04
(231)
238.03
(237)
(242)
(243)
Pa
91
U
92
Np
93
Pu
94
Am Cm Bk Cf
95 96 97 98
Ac Th
89 90
Silver
Cadmium
Sb
51
Te
52
Bismuth Polonium
Astatine
Radon
200.98
(210)
(210)
(222)
Bi Po
83 84
At
85
Rn
86
104 105 106 107 108 109 110 111 112 113 114 115 116 117 118
140.91
La
57
Indium
107.87 112.40
151.96 157.25 158.92 162.50
Erbium
164.93 167.26
Thulium
Ytterbium
Lutetium
168.93
173.04
174.97
Yb
70
Lu
71
Eu Gd Tb Dy Ho Er Tm
63 64 65 66 67 68 69
Neptunium Plutonium Americium
Curium Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium
(247)
(249)
228
(251)
(254)
(253)
(256)
(254)
(257)
Es Fm Md No Lr
99 100 101 102 103
APPENDIX F: SAMPLE UNITS FOR SOME COMMON VARIABLES
This certainly doesn’t cover all the possible ways each of these variables can be described, but
we’ve tried to include the units you will most commonly see.
ACCELERATION (a):
m N

,
s 2 kg
FORCE (F): newtons  N 
ft
lb

2
slug
s
kg  m
slug  ft
, pounds  lb 
2
s
s2
kg  m 2
slug  ft 2
,
ft

lb

, in  lb
s2
s2
N
J
lb
lb
PRESSURE (P): pascal  Pa  2  3 , psi  2 , psf  2
m
m
in
ft
TORQUE (): N  m 
kg  m
slug  ft 2
,
,
ft

lb

s2
s2
calories =cal, British Thermal Units = BTU
2
WORK (W) = ENERGY (E): joules  J  N  m 
CURRENT (I): amperes  amps  A 
FREQUENCY (f):
C
s
cycles
 hertz  Hz
s
VOLUME FLOW RATE (QV):
THERMAL RATE (RT):
m 3 cm 3  ft 3 gal
,
, ,
,
s
s hr hr min
cal BTU
,
s
hr
DRAG RESISTANCE (RD):
N
lb
,
m / s ft / s
FLUID RESISTANCE (RF):
Pa
N / m 2 Pa psi psf psi

,
,
,
m 3 / s m 3 / s  / hr gpm gpm ft 3 / hr
ELECTRICAL RESISTANCE (R, or RE):
ohms   
V Js

A C2
C
F
,
THERMAL RESISTANCE (RT):
cal / sec BTU / hr
ANGULAR VELOCITY ():
radians rad rev
rev

,
 rpm ,
s
s min
s
229
J
C2
C2  s 2
CAPACITANCE (C): farad  F  2 

J
V
kg  m 2
INDUCTANCE (L):
henries H 
LINEAR MOMENTUM (p):
J
J  s 2 V  s kg  m 2



A
A2
C2
C2
kg  m
 N  s,
s
ANGULAR MOMENTUM (L):
slug  ft
 lb  s
s
kg  m 2
 N  m  s  J  s,
s
slug  ft 2
 ft  lb  s
s
POWER (P):
J N  m kg  m 2


,
s
s
s3
cal
BTU
horsepower  hp,
,
s
hr
watts  W 
IRRADIANCE ():
W mW hp
,
,
m 2 cm 2 ft 2
230
ft  lb slug  ft 2

,
s
s3
APPENDIX G: GLOSSARY OF TERMS
Acceleration (a) is the rate at which velocity changes with time. It’s typically measured in units
of m/s2 (or sometimes N/kg, which is the same thing) or ft/s2. The gravitational
acceleration (g) at sea level is 9.8 m/s2 or 32 ft/s2.
Angular acceleration () is the rate at which an object changes its angular velocity ().
Typical units are rad/sec2 or rev/sec2.
Angular Impulse is the product of torque and the time over which it is applied. It has units of
kg·m2/s = N·m·s.
Angular momentum (L) describes the momentum of a rotating object. It is the product of the
moment of intertia and the angular velocity and has units of kg·m2/s = N·m·s.
Angular velocity ()is the rate at which an object changes its angular position. Typical units
are rad/sec, rev/sec and rev/min (rpm).
British Thermal Unit (BTU) is an English unit of thermal energy. More specifically, it is the
amount of thermal energy required to raise the temperature of one pound of liquid water
(at one atmosphere pressure) by 1 Fo.
Buoyant Force (FB) is the upward force on an object due to the weight of the fluid it displaces.
calorie (cal) is a metric unit of thermal energy. More specifically, one calorie is the amount of
thermal energy required to raise the temperature of one gram of liquid water (at one
atmosphere pressure) by 1 Co.
Capacitance (C) describes a capacitor’s ability to hold charge for a given voltage difference. It
is in units of charge per potential difference: 1 farad = 1 F = 1coulomb/volt = 1 C/V = 1
C2/J = 1 C2·s2/kg·m2.
Charge (q) is that mysterious carrier of the electromagnetic force. The units of charge are
coulombs (1 coulomb = 1 C).
Coefficient of friction is the ratio of the frictional force to the normal force. It has no units. The
frictional force for the static coefficient of friction (s) is the maximum force that can be
applied (horizontal to the surface) without having the object slide. The frictional force
for the kinetic coefficient of friction (k) is the surface frictional force resisting an object
in motion. Both s and k are dimensionless.
Conductivity () is a measure of a substance’s ability to conduct charge or thermal energy.
Conductors are materials that readily transmit charge or thermal energy.
Cooling Rate (R) is the rate at which a temperature change occurs. Typical units are C°/min,
C°/sec, F°/min, F°/sec, etc.
Current (I) in an electrical system is the rate at which charge moves. 1 ampere = 1 amp = 1 A =
1 coulomb/second = 1 C/s.
Dynamic pressure (1/2v2) is the pressure due to the motion of a fluid.
Energy (E) is a property of the universe. Doing work means using energy. Units are typically
joules (1 J = 1 N·m = 1 kg·m2/s2) or foot-pounds (1 ft·lb = 1 slug·ft2/s2).
Frequency (f) is the rate at which a periodic (regularly repeating) event occurs. Its units are
generally cycles per second, also known as hertz (Hz). It is the inverse of period.
Force (F) is the product of mass (m) and acceleration (a), and so are its units: 1 newton = 1 N =
1 kg·m/s2, and 1 pound = 1 lb = 1 slug·ft/s2. If you want to change the motion of an
object, you have to apply a force (F) on it.
Gravitational acceleration (g) is the acceleration due to the presence of a gravitational field. At
sea level, g = 9.8 m/s2 or 32 ft/s2.
Heat (H) is more properly described as thermal energy. It comes in units of calories and British
thermal units (Btu’s).
Heat Flow Rate (QH) or heat transfer rate is the rate at which thermal energy is transferred.
Typical units are cal/sec or Btu/sec or even watts (1 watt = 1 W = 1 J/s = 1 kg·m2/s3).
231
Impulse is the change in momentum (p) of an object that is produced by an applied force
over a given time interval (p=F·t).
Inductance (L) is a measure of an inductor’s ability to store energy in its magnetic field, as a
function of the current flowing through it. It has units of henries (1 henry = 1 H = 1
V·s/A = 1 J/A2 = 1 kg·m2/C2).
Insulators are materials that inhibit the flow of charge or thermal energy.
Kinetic Energy (Ek) is energy of motion and has the units of energy (work).
Mass (m) is a measure of the quantity of matter in an object. It’s measured in units of kilograms
(kg) or slugs.
Mass density (), or just density, is mass per unit volume. Typical units are kg/m3 and g/cm3.
Moment of Inertia (I) describes an object’s resistance to rotational motion based on its mass,
the distribution of its mass from the axis of rotation and the axis of rotation. Its units are
kgm2 or slugft2.
Momentum (p) is the product of mass (m) and velocity (v). Its units are kg·m/s = N·s or
slug·ft/s = lb·s.
Period (T) is the time for one complete cycle of periodic (regularly repeating) motion. Its units
are those of time (or time per cycle). It is the inverse of frequency.
Potential Energy (Ep) is stored energy and has the units of energy (work).
Power (P) is the rate at which energy is used, or the rate at which work is done. Power is
measured in watts (1 watt = 1 W = 1 J/s = 1 kg·m2/s3), in foot-pounds per second (ft·lb/s),
in horsepower (hp), in calories per second (cal/s), or in Btu/s.
Pressure (P) is force per unit area (F/A). Units are pascals (1 Pa = 1 N/m2), lb/ft2, lb/in2, in.
Hg, in. H2O, mm Hg., mm H2O, bars and millibars, to name a few.
Resistance (R) in electrical system is the resistance to current flow in a circuit. The units are
ohms: 1 ohm = 1  = 1 joule-second/(coulomb squared) = 1 J·s/C2.
Resistivity () is a measure of a substance’s ability to restrict charge or thermal energy. Its units
are typically ·cm.
Reynolds Number (Re) is used to determine the level of turbulence in a fluid system. It is
dimensionless.
Scalar is a quantity with magnitude but no direction, like “24 oC”, or “32 years old”.
Specific Heat (c) is a measure of the amount of energy required to change the temperature of a
given amount of a certain substance. It is given in units of cal/g·C° or Btu/lb·F°.
Temperature (T) is a statistical measure of a substance’s kinetic energy.
Thermal Resistance (RT) is a substance’s (or an object’s) resistance to heat conduction.
Typical units are F°/(Btu/hr) and C°/(cal/hr).
Torque () is the result of a force (F) applied perpendicular to a moment arm (). Typical units
are N·m, ft·lb and in·lb.
Vector is a quantity with magnitude and direction like “4 newtons at 60o” or “6 miles east”.
Velocity (v) is the change in distance over time. Typical units are m/s, ft/s, km/hr and mph.
Voltage (V) or potential difference is the energy per charge in an electrical system. 1 volt = 1
V = 1 joule/coulomb = 1 J/C.
Volume Flow Rate (QV) is the rate at which volume flows. It typically has units of ft3/sec,
ft3/min, gal/min (=gpm), gal/hr, liters/min, m3/min, m3/sec, etc.
Weight (w) is the force due to an object¹s mass and the local gravitational acceleration (g).
Units are the units of force.
Weight density (w) is weight per unit volume. Typical units are lb/ft3 and lb/gal.
Work (W) is the product of an applied force and the distance over which it is applied. Units are
those of energy.
232
Appendix H: The Greek Alphabet
Capital Low-case Greek Name English
Alpha
a
Beta
b
Gamma
g
Delta
d
Epsilon
e
Zeta
z
Eta
h
Theta
th
Iota
i
Kappa
k
Lambda
l
Mu
m
Nu
n
Xi
x
Omicron
o
Pi
p
Rho
r
Sigma
s
Tau
t
Upsilon
u
Phi
ph
Chi
ch
Psi
ps
Omega
o
233