02 chapter 11 thermodynamics
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Engineering 112Foundations of Engineering
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Student Information Sheet
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Engineering Disciplines
Electrical Engineering Civil Engineering Mechanical Engineering Industrial Engineering Aerospace Engineering Chemical Engineering
Biomedical Engineering Materials Engineering Agricultural Engineering Nuclear Engineering Architectural Engineering Petroleum Engineering Engineering Technology
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Course Syllabus PurposeMaterialExamsGradingCourse Policies
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Objectives of ENGR 112
Develop a better understanding of enginesBecome a better problem solver Develop a mastery of unit analysisImprove your mathematics skills Prepare you for statics and dynamicsDevelop teaming skills
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Course Calendar
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A Brief History of EGR111/112
These courses were added to the curriculum at TAMU in the early 1990’s.
12 disciplines require these courses.The courses were first taught at SFA
starting in the Fall of 2002.They are part of an articulation agreement
with TAMU.They also transfer to other universities.
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Course Description
PHY108 Introduction to PHY/EGR
EGR111 Foundations I
EGR112 Foundations II
EGR215 Electrical Engineering
EGR343 Digital Systems
EGR250 Engineering Statics
EGR321 Engineering Dynamics
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CoursePre-EGR
DUAL Minor
PHY108
EGR111
EGR112
EGR215 ~ ~
EGR342 ~ ~
PHY250 ~
PHY321 ~
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Teaming Expectations
Many of the activities in ENGR 112 require collaboration with other class members
Each student will be assigned to a teamAll students will receive team training
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Before Wednesday…
Get a Note Book and Text BookDouble Check you Schedule
4th Class Day12th Class DayMid-Semester
Complete Problems 1 – 5 on HW1
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Can you boil water at room temperature?
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How can you design a room that is completely silent?
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Thermodynamics
Chapter 11
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Thermodynamics
Developed during the 1800’s to explain how steam engines converted heat into work.
Thought Questions:Is heat just like light and sound?Is there a “speed of heat”?
Answer: Not really.
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11.1 Forces of Nature
Gravity ForceElectromagnetic ForceStrong ForceWeak Force
Nuclear Forces
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Chapter 11 - Thermodynamics
11.1 - Forces of Nature
11.2 - Structure of Matter
11.3 - Temperature
11.4 - Pressure
11.5 - Density
11.6 - States of Matter
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11.2 Structure of Matter
ProtonsAtomic Number - number of protons
Neutronsnuclear glue
ElectronsValence Electrons - those far from the nucleus
Atoms, Molecules, and a LatticeAmorphous - random arrangement of atomsCrystal - atoms are ordered in a lattice
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Which is colder?
Metal or Wood?
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11.3 Temperature
Measured in Fahrenheit, Celsius, and Kelvin
Rapidly moving molecules have a high temperature
Slowly moving molecules have a low temperature
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Cool Hot
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What is “absolute zero”?
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Temperature Scales
Fahrenheit Celsius Kelvin
Boiling Pointof Water
Freezing Pointof Water
Absolute Zero
212F
32F
-459F
100C
0C
-273C
373 K
273 K
0 K
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11.4 Pressure
Pressure - force per unit area It has units of N/m2 or Pascals (Pa)
A
FP
F
A
Impact Weight
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Pressure
What are the possible units for pressure?N/m2
Pascal 1 Pa = 1 N/m2
atm 1 atm = 1 × 105 Papsi 1 psi = 1 lb/inch2
mm Hg 1 atm = 760 mm Hg
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11.5 Density
Density - mass per unit volumeIt has units of g/cm3
V
M
High densityLow density
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11.6 States of Matter
Solid Liquid
Gas Plasma
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State of Matter Definitions
Phase DiagramPlot of Pressure versus Temperature
Triple PointA point on the phase diagram at which all
three phases exist (solid, liquid and gas)
Critical PointA point on the phase diagram at which the
density of the liquid a vapor phases are the same
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Figure 11.8 - Phase Diagram
Plasma
Gas
Vapor
Liquid
Solid
Ttriple Tcritical
Ptriple
Pcritical
Pressure
Temperature
Critical Point
TriplePoint
Boiling
Condensation
Sublimation
Melting
Freezing
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Questions
Is it possible to boil water at room temperature? Answer: Yes. How?
Is it possible to freeze water at room temperature? Answer: Maybe. How?
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Gas Laws
Perfect (ideal) GasesBoyle’s LawCharles’ LawGay-Lussac’s LawMole Proportionality Law
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Boyle’s Law
2
12
1 V
V
P
P
T = const n = const
P1
V1
P2
V2
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Charles’ Law
1
2
1
2
T
T
V
VT1
V1
T2
V2
P = const n = const
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Gay-Lussac’s Law
1
2
1
2
T
T
P
PT1
P1
T2
P2
V = const n = const
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Mole Proportionality Law
1
2
1
2
n
n
V
V
T = const P = const
n1
V1
n2
V2
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Thermodynamics
Chapter 11Homework 1
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Boyle’s Law
2
12
1 V
V
P
P
T = const n = const
P1
V1
P2
V2
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Charles’ Law
1
2
1
2
T
T
V
VT1
V1
T2
V2
P = const n = const
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Gay-Lussac’s Law
1
2
1
2
T
T
P
PT1
P1
T2
P2
V = const n = const
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Mole Proportionality Law
1
2
1
2
n
n
V
V
T = const P = const
n1
V1
n2
V2
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Perfect Gas Law
The physical observations described by the gas laws are summarized by the perfect gas law (a.k.a. ideal gas law)
PV = nRTP = absolute pressureV = volumen = number of molesR = universal gas constantT = absolute temperature
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Table 11.3: Values for R
mol·Katm·L
mol·KPa·m3
08205.0
314.8
mol·K
cal
mol·K
J
1.987
314.8
Work Problem 11.8
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Thermodynamics
Chapter 11Movie R.A.T.
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RAT Movies
For the movies that follow, identify the gas law as a team.
Only the recorder should do the writing.
Turn in the team’s work with the team name at the top of the page.
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Balloon Example (Handout)
A balloon is filled with air to a pressure of 1.1 atm.
The filled balloon has a diameter of 0.3 m. A diver takes the balloon underwater to a
depth where the pressure in the balloon is 2.3 atm.
If the temperature of the balloon does not change, what is the new diameter of the balloon? Use three significant figures.
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Volumes?
CubeV=a3
SphereV=4/3 r3
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33
2
112
2
131
32
31
3
11
32
3
22
2
112
2
1
1
2
atm 3.2
atm 1.1 m 3.0
23
4
23
4
P
PDD
P
PkDkD
kDD
V
kDD
V
P
PVV
P
P
V
V
= 0.235 m
Solution
P1 = 1.1 atmD1 = 0.3 m
P2 = 2.3 atmD2 = ?
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Work
Work = Force Distance W = F x
The unit for work is the Newton-meter which is also called a Joule.
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Joule’s ExperimentJoule showed that mechanical energy could beconverted into heat energy.
F
M
xH2O
T
W = Fx
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Heat Capacity Defined
Q - heat in Joules or caloriesm - mass in kilogramsT - change the temperature in KelvinC has units of J/kg K or kcal/kg K1 calorie = 4.184 Joules
Tm
QC
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F
m
xH2O
T
Tm
QC
W = Fx 1 kcal= 4184 J
Problem 11.9
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Heat Capacity
An increase in internal energy causes a rise in the temperature of the medium.
Different mediums require different amounts of energy to produce a given temperature change.
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Myth Busters - Cold Coke
Do you burn more calories drinking a warm or cool drink?
How many calories do you burn drinking a cold Coke?Assume that a coke is 335ml and is chilled to
35F and is about the same density and heat capacity as water. The density of water is 1g/cm3.
1 kcal=4184 J 1ml=1cm3
The heat capacity of water is 1 calorie per gram per degree Celsius (1 cal/g-°C).
TC = (5/9)*(TF-32)
Tm
QC
http://en.wikipedia.org/wiki/Calorie
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Thermodynamics
Chapter 11
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11.11 Energy
Energy is the ability to do work.It has units of Joules.It is a “Unit of Exchange”.Example
1 car = $20k1 house = $100k5 cars = 1 house =
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11.11 Energy Equivalents
What is the case for nuclear power?1 kg coal » 42,000,000 joules1 kg uranium » 82,000,000,000,000
joules1 kg uranium » 2,000,000 kg coal!!
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11.11.3 Energy Flow
Heat is the energy flow resulting from a temperature difference.
Note: Heat and temperature are not the same.
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Heat Flow
T = 100oC
T = 0oC
Temperature Profile in Rod
HeatVibrating copper atom
Copper rod
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11.12 Reversibility
Reversibility is the ability to run a process back and forth infinitely without losses.
Reversible Process Example: Perfect Pendulum
Irreversible Process Example: Dropping a ball of clay
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Reversibility Movies of reversible phenomena
appear the same when played forward and backward.
IrreversibilitiesThe opposite is true.
“Movie Making”
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Reversible Process
Examples: Perfect PendulumMass on a SpringDropping a perfectly elastic ballPerpetual motion machinesMore?
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Irreversible Processes
Examples:Dropping a ball of clayHammering a nailApplying the brakes to your carBreaking a glassMore?
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Example: Popping a Balloon
Not reversible unless energy is expended
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Sources of Irreversibilities
Friction (force drops)Voltage dropsPressure dropsTemperature dropsConcentration drops
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First Law of Thermodynamicsenergy can neither be created
nor destroyed
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Second Law of Thermodynamicsnaturally occurring processes are
directionalthese processes are naturally
irreversible
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Third Law of Thermodynamicsa temperature of absolute zero is
not possible
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Heat into Work
Thot TcoldHeat
Engine
W
QhotQcold
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Carnot Equation: Efficiency
The maximum work that can be done by a heat engine is governed by:
hot
cold
hot
max
T
T
Q
WEfficiency 1
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What is the maximum efficiency that a heat engine can have using steam and an ice bath?
Team Exercise (3 minutes)
Thot TcoldHeat
Engine
W
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Work into Heat
Although there are limits on the amount of heat converted to work, work may be converted to heat with 100% efficiency.
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Chapter 12
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Heat Capacity for Constant Volume Processes (Cv)
Heat is added to a substance of mass m in a fixed volume enclosure, which causes a change in internal energy, U. Thus,
Q = U2 - U1 = U = m Cv TThe v subscript implies constant volume
Heat, Qaddedm m
Tinsulation
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Heat Capacity for Constant Pressure Processes (Cp)
Heat is added to a substance of mass m held at a fixed pressure, which causes a change in internal energy, U, AND some PV work.
Heat, Qadded
T
m m
x
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Cp Defined
Thus,Q = U + PV = H = m Cp T
The p subscript implies constant pressure
H, enthalpy. is defined as U + PV, so H = (U+PV) = U + VP + PV = U + PV
Experimentally, it is easier to add heat at constant pressure than constant volume, thus you will typically see tables reporting Cp for various materials (Table 21.1 in your text).
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Individual Exercises (5 min.)
1. Calculate the change in enthalpy per unit lbm of nitrogen gas as its temperature decreases from 1000 oR to 700 oR.
2. Two kg of water (Cv=4.2 kJ/kg K) is heated by 200 BTU of energy. What is the change in temperature in K? In oF?
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Solution
1. From table 21.2, Cp for N2 = 0.249 BTU/lbmoF.
Note that since oR = oF + 459.67, then T oR = T oF, so
2.
F2.45))K)( 1.52((
K 1.25
)2.4( kg 2
)055.1(BTU 200
Kin change 1Fin change 1.8
K kgkJ
BTUkJ
vmC
QT
Recall, we are referring to a temperature CHANGE
m
mp
lb
BTU 49.74
)F 300(Flb
BTU249.0TC
m
H
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Homework
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http://demoroom.physics.ncsu.edu/html/demos/88.html
Exercise
A stick man is covered with marshmallows and placed in a sealed jar.
What will happen to the marshmallow man when the jar is evacuated? Why?
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Solution
Click to activate, then click play Suggestion: view at 200%
marshmallow.mov
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Other Homework Questions
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What’s next?
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Example Problem
A cube of aluminum measures 20 cm on a side sits on a table.
Calculate the pressure (N/m2) at the interface.
Note: Densities may be found in your text.
A
FP
V
M
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L = 0.2 m
L = 0.2 m
L = 0.2 mA
FP
mgF 3LVm
2LA
Solution
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Heat/Work Conversions
Heat can be converted to work using heat enginesJet engines (planes)steam engines (trains)internal combustion engines
(automobiles)
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Team Exercise (2 minutes)
On the front of the page write down 2 benefits of working in a team.
On the back write down 1 obstacle that we must overcome to work in engineering teams.
You have two minutes…
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Why Teamwork
Working in groups enhances activities in active/collaborative learning
Generate more ideas for solutionsDivision of laborBecause that’s the way the real world
works!!Industry values teaming skills
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Why Active/Collaborative Learning
Activecountless studies have shown
improvement in:short-term retention of material,long-term retention of material,ability to apply material to new situations
Collaborativeby not wasting time on things you already
know we can make the best use of class time
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Teamwork Obstacles
What are some potential problems with teamwork? “I’m doing all of the work.”
Solution: It is part of your team duties to include everyone in a team project.
“I feel like I’m teaching my teammates.” Exactly. By explaining difficult concepts to your team members
your grasp of difficult concepts can improve.
“What if I don’t get along with my teammates.” Solution: This is a problem that all workers have at some point. The team may visit with the instructor during office hours to iron out
differences.
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Project One
The Rubber Band Heat Engine
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Example Problemsfrom Homework
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Let’s take notes…
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Boyle’s Law
2
12
1 V
V
P
P
T = const n = const
P1
V1
P2
V2
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Charles’ Law
1
2
1
2
T
T
V
V
T1
V1
T2
V2
P = const n = const
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Gay-Lussac’s Law
1
2
1
2
T
T
P
P
T1
P1
T2
P2
V = const n = const
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Mole Proportionality Law
1
2
1
2
n
n
V
V
T = const P = const
n1
V1
n2
V2
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Problems
Homework 1111213
In-class AssignmentProblem 1
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Problems
Homework 114
In-class AssignmentProblem 2